Rd Sharma XII Vol 2 2018 Solutions for Class 12 Science Math Chapter 2 Areas Of Bounded Regions are provided here with simple step-by-step explanations. These solutions for Areas Of Bounded Regions are extremely popular among Class 12 Science students for Math Areas Of Bounded Regions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2018 Book of Class 12 Science Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2018 Solutions. All Rd Sharma XII Vol 2 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 21.14:

Question 1:

Using integration, find the area of the region bounded between the line x = 2 and the parabola y2 = 8x.

Answer:



y2=8x  represents a parabola with vertex at origin and axis of symmetry along the +ve direction of x-axisx=2 is line parallel to y-axisLet (x, y)  be a given point on the parabola , y2=8xSince parabola   y2=8x   is symmetric about x-axis ,Required area =2 area OCAO On slicing the area above x-axis into vertical strips of length =y   and width =dx area of rectangular strip=y dxThe approximating rectangle moves between x=0 and x=2. So, area A=2 area OCAOA=202y dx=202y dx           as y>0  A=2028x dx A=2×2022x dx= 4202x dxA= 42 x323202=832 232-0 =83×22=323  sq. units Area A=323 sq. units

Page No 21.14:

Question 2:

Using integration, find the area of the region bounded by the line y − 1 = x, the x − axis and the ordinates x = −2 and x = 3.

Answer:




 y-1 =x  is a straight line cutting x-axis at D(-1, 0) and y-axis at A( 0, 1) And, x=-2 and x=3 are straight  lines parallel to y-axis Also, since y =-1+x,       for x-1And             y = 1+x,                x>-1 Area of region bound by line y-1=x , x axis and the ordinates x=-2 and x=3 isArea A = area FED+area DOA +area OABCA=-23y dxA=-2-1y dx   +-10y dx  +03y dxA=-2-1-1+x dx +-101+x dx+031+x dxA=-x+x22-2-1+x+x22-10+x+x2203A=1-12-2+42+0+1-12+3+92A=-1+32+1-12+3+92=3+3-1+92=3+112=172  sq. unitsArea of the bound region =172  sq. units



Page No 21.15:

Question 3:

Find the area of the region bounded by the parabola y2 = 4ax and the line x = a.

Page No 21.15:

Question 4:

Find the area lying above the x-axis and under the parabola y = 4xx2.

Answer:




The equation y=4x-x2 represents a parabola opening downwards and cutting the x axis at O(0, 0) and B(4, 0)Slicing the region above x axis in vertical strips of length= y and width =dx , area of corresponding rectangle is = y dxSince the corresponding rectangle can move from x=0 to x=4,Required area OABO isA=04y dx= 04y dx                    As, y>0 for 0 x 4  y =y A=044x-x2 dx   A=4x22-x3304 A=32-643 A=323 square units

Page No 21.15:

Question 5:

Draw a rough sketch to indicate the region bounded between the curve y2 = 4x and the line x = 3. Also, find the area of this region.

Answer:



y2=4x  represents a parabola with vertex at (0, 0) and axis of symmetry along the +ve direction of x axis x=3 is a line parallel to y axis and cutting x axis at (3, 0)Since  y2=4x  is symmetrical about x axis , Required area A=OACO=2× area OABOSlicing the area above x axis into vertical strips of length=y and width = dx    Area of corresponding rectangle= y dxThe corresponding rectangle moves from x=0 to x=3A=2× area OABOA=203 y dx=  203 y dx                   As, y=y, y>0A=  2034x dx A=403x dx A=4x323203= 83x3203=83×33=83  sq. unitsArea of region bound by curve y2=4x and x=3  is  83  sq. units

                   

         

     

Page No 21.15:

Question 6:

Make a rough sketch of the graph of the function y = 4 − x2, 0 ≤ x ≤ 2 and determine the area enclosed by the curve, the x-axis and the lines x = 0 and x = 2.

Answer:




y=4-x2, 0x2 represents a half parabola with vetex at (4, 0) x=2 represents a line parallel to y-axis and cutting x-axis at (2, 0) In quadrant OABO, consider a vertical strip of length= y, width=dxArea of approximating rectangle= y dx  The approximating rectangle  moves from x=0 to x=2A=Area OABO  =02 y dx A=02 y dx                            As, y>o, y =yA=024-x2 dx   A=4x-x3302A=8-83A=163 sq. unitsThe area enclosed by the curve and x-axis and given lines =163 sq. units

Page No 21.15:

Question 7:

Sketch the graph of y = x+1 in [0, 4] and determine the area of the region enclosed by the curve, the x-axis and the lines x = 0, x = 4.

Answer:



y=x+1  in 0, 4 represents a curve which is part of a parabola x=4 represents a line parallel to y-axis and cutting x-axis at (4, 0)Enclosed area bound by the curve and lines x=0 and x=4 is OABCOConsider a vertical strip  of lenght =y and width=dxArea of approximating rectangle= y dxThe approximating rectangle moves from x=0 to x=4A=Area OABCO=04y dxA=04y dx                  y>0y =yA=04x+1 dxA=04x+112 dxA=x+1323204A=23532-1 sq. units Enclosed area between the curve and given lines =23532-1  sq. units

Page No 21.15:

Question 8:

Find the area under the curve y = 6x+4 above x-axis from x = 0 to x = 2. Draw a sketch of curve also.

Answer:




y=6x+4  represents a parabola, with vertex V(-23, 0) and symmetrical about x-axis x=0 is the y-axis . The curve cuts it at A(0, 2 ) and A'(0, -2)x=2 is a line parallel to y-axis, cutting the x-axis at  C(2, 0)The enclosed area of the curve between x=0 and x=2 and above x-axis =area OABCConsider, a vertical strip of length =y  and width =dx Area of approximating rectangle =y dxThe approximating rectangle moves from x=0  to x=2  area OABC =02y dxA=02y dx                As, y>0 , y =yA=026x+4 dx  A=026x+412 dxA=166x+4323202A=2181632-432A=21843-23A=21864-8 =218×56 =569 sq. unitsEnclosed area =569 sq. units

Page No 21.15:

Question 9:

Draw the rough sketch of y2 + 1 = x, x ≤ 2. Find the area enclosed by the curve and the line x = 2.

Answer:


y2+1=x, x2  is a parabola with vertex (1, 0) and symmetrical about +ve side of x-axis x=2 is a line parallel to y-axis and cutting x-axis at (2, 0)Consider, a vertical section of height= y  and width =dx in the first quadrant area of corresponding rectangle =  y dxSince, the corresponding rectangle is moving from x=1 to x=2 and the curve being symmetricalA=area ABCA=2×area ABDAA=212y dx A=212y dx                          y=y as y>0A=212x-1 dx                y2+1=x  y=x-1 A=2x-1323212  A=43132-0A=43  sq. unitsEnclosed area=43 sq. units

Page No 21.15:

Question 10:

Draw a rough sketch of the graph of the curve x24+y29=1 and evaluate the area of the region under the curve and above the x-axis.

Answer:


Since in the given equation  x24+y29 =1, all the powers of both x and y are even, the curve is symmetrical about both the axis .Area encloed by the curve and above x axis = area A'BA =2×area enclosed by ellipse and x-axis in first quadrant(2, 0 ), (-2, 0) are the points of intersection of curve and x-axis(0, 3), (0, -3) are the points of intersection of curve and y-axisSlicing the area in the first quadrant into vertical stripes of height =y  and width =dxArea of approximating  rectangle =y dxApproximating  rectangle can move between x=0 and x=2 A=Area of  enclosed curve above x-axis =202y dxA=202y dxA=202324-x2 dxA=3024-x2 dxA=312x 4-x2 +12 4 sin-1x02=30+12×4 sin-11=3×12×4×π2= 3π sq. unitsArea of enclosed region above x-axis = 3π sq. units

Page No 21.15:

Question 11:

Sketch the region {(x, y) : 9x2 + 4y2 = 36} and find the area of the region enclosed by it, using integration.

Answer:



We have,9x2+4y2=36     .....14y2=36-9x2y2=944-x2y=324-x2     .....2From 1 we getx24+y29 =1Since in the given equation  x24+y29 =1, all the powers of both x and y are even, the curve is symmetrical about both the axes.Area enclosed by the curve and above x axis = 4×area enclosed by ellipse and x-axis in first quadrant(2, 0 ), (-2, 0) are the points of intersection of curve and x-axis(0, 3), (0, -3) are the points of intersection of curve and y-axisSlicing the area in the first quadrant into vertical stripes of height =y  and width =dxArea of approximating  rectangle =y dxApproximating  rectangle can move between x=0 and x=2A=Area of  enclosed curve=402y dxA=402324-x2dx    From 2=4×32024-x2dx=60222-x2dx=6x222-x2+1222sin-1x202=60+124sin-11=612×4π2A=6π sq units

Page No 21.15:

Question 12:

Draw a rough sketch of the graph of the function y = 21-x2, x ∈ [0, 1] and evaluate the area enclosed between the curve and the x-axis.

Answer:



We have,y=21-x2y2=1-x2y24=1-x2x2+y24=1x21+y24=1Since in the given equation  x21+y24 =1, all the powers of both x and y are even, the curve is symmetrical about both the axes .Required area=area enclosed by ellipse and x-axis in first quadrant(1, 0 ), (-1, 0) are the points of intersection of curve and x-axis(0, 2), (0, -2) are the points of intersection of curve and y-axisSlicing the area in the first quadrant into vertical stripes of height =y  and width =dxArea of approximating  rectangle =y dxApproximating  rectangle can move between x=0 and x=1 A=Area of  enclosed curve above x-axis =01y dxA=0121-x2 dx=2011-x2 dx=212x1-x2+12sin-1x01=212sin-11=212π2-0A=π2 sq.units

Page No 21.15:

Question 13:

Determine the area under the curve y = a2-x2 included between the lines x = 0 and x = a.

Answer:



We have,y=a2-x2y2=a2-x2x2+y2=a2Since in the given equation  x2+y2=a2, all the powers of both x and y are even, the curve is symmetrical about both the axis .Required area = area enclosed by circle in first quadrant(a, 0 ), (-a, 0) are the points of intersection of curve and x-axis(0, a), (0, -a) are the points of intersection of curve and y-axisSlicing the area in the first quadrant into vertical stripes of height =y  and width =dxArea of approximating  rectangle =y dxApproximating  rectangle can move between x=0 and x=aA=Area of  enclosed curve in first quadrant =0ay dxA=0aa2-x2 dx=12xa2-x2+12a2sin-1xa0a=12a2sin-11=12a2π2 =a2π4 sq units

Page No 21.15:

Question 14:

Using integration, find the area of the region bounded by the line 2y = 5x + 7, x-axis and the lines x = 2 and x = 8.

Answer:



We have,
Straight line 2y = 5x + 7 intersect x-axis and y-axis at ( −1.4, 0) and (0, 3.5) respectively.
Also x = 2 and x = 8 are straight lines as shown in the figure.
The shaded region is our required region whose area has to be found.
When we slice the shaded region into vertical strips, we find that each vertical strip has its lower end on x-axis and upper end on the line
2y = 5x + 7
So, approximating rectangle shown in figure has length = y and width = dx and area = y dx.
The approximating rectangle can move from x = 2 to x = 8.
So, required is given by,

A=28y dx=285x+72 dx=1228(5x+7) dx=1252x2+7x28=1252×64+56-52×4-14=12×192=96 sq units

Page No 21.15:

Question 15:

Using definite integrals, find the area of the circle x2 + y2 = a2.

Answer:


Area of the circle x2 + y2 = a2 will be the 4 times the area enclosed between x = 0 and x = a in the first quadrant which is shaded.

A=40ay dx=40aa2-x2 dx=412xa2-x2+12a2sin-1xa0a=40+12a2sin-11=412a2π2           sin-11=π2=a2π sq units

Page No 21.15:

Question 16:

Using integration, find the area of the region bounded by the following curves, after making a rough sketch: y = 1 + | x + 1 |, x = −2, x = 3, y = 0.

Answer:



We have,

y = 1 + | x + 1 | intersect x = − 2 and at ( −2, 2) and x = 3 at (3, 5).
And y = 0 is the x-axis.
The shaded region is our required region whose area has to be found

y=1+x+1=1-x+1    x<-11+x+1      x1=-x       x<-1x+2      x1

Let the required area be A. Since limits on x are given, we use horizontal strips to find the area:
A=-23ydx=-2-1ydx+-13ydx=-2-1-x dx+-13x+2dx=-x22-2-1+x22+2x-13=-12-42+92+6-12+2=32+8+82=32+8+4=272 sq. units     

Page No 21.15:

Question 17:

Sketch the graph y = | x − 5 |. Evaluate 01x-5 dx. What does this value of the integral represent on the graph.

Answer:

We have,
y = | x − 5 | intersect x = 0 and x = 1 at (0, 5) and (1, 4)
Now,
y=x-5=-x-5     For all x0, 1

Integration represents the area enclosed by the graph from x = 0 to x = 1
A=01ydx=01x-5 dx=01-x-5 dx=-01x-5 dx=-x22-5x01=-12-5-0-0=--92=92 sq. units


Page No 21.15:

Question 18:

Sketch the graph y = | x + 3 |. Evaluate -60x+3 dx. What does this integral represent on the graph?

Answer:



We have,
y = | x + 3 | intersect x = 0 and x = −6 at (0, 3) and (−6, 3)
Now,
y=x+3=x+3     For all x>-3-x+3     For all x<-3


Integral represents the area enclosed between x = 6 and x = 0
A=-60y dx=-6-3y dx+-30y dx=-6-3-x+3 dx+-30x+3 dx=-x22+3x-6-3+x22+3x-30=-92-9-362+18+0+0-92+9=-92+9+362-18-92+9=9 sq. units

Page No 21.15:

Question 19:

Sketch the graph y = | x + 1 |. Evaluate -42x+1 dx. What does the value of this integral represent on the graph?

Answer:

We have,
y = | x + 1 | intersect x = −4 and x = 2 at (−4, 3) and (2, 3) respectively.
Now,
y=x+1=x+1     For all x>-1-x+1     For all x<-1


Integral represents the area enclosed between x = −4 and x = 2
A=-42ydx=-4-1ydx+-12ydx=-4-1-x+1dx+-12x+1dx=-x22+x-4-1+x22+x-12=-12-1-162+4+42+2-12+1=-3-152+5-12=-3+152+5-12=9 sq. units

Page No 21.15:

Question 20:

Find the area of the region bounded by the curve xy − 3x − 2y − 10 = 0, x-axis and the lines x = 3, x = 4.

Answer:

We have,
xy-3x-2y-10=0xy-2y=3x+10yx-2=3x+10y=3x+10x-2
Let A represent the required area:
A=34ydx=343x+10x-2dx=343x-6+16x-2dx=343+16x-2dx=3x+16 log x-234=12+16 log 2 -9 -16 log 1=3+16 log 2 sq. units

Page No 21.15:

Question 21:

Draw a rough sketch of the curve y = π2+2 sin2 x and find the area between x-axis, the curve and the ordinates x = 0, x = π.

Answer:



 

x 0 π6 π2 5π6 π
sin x 0 12 1 12 0
y=π2+2sin2x 1.57 2.07 3.57 2.07 1.57


y=π 2+2 sin2 x is an arc cutting y-axis at (1.57, 0 ) and x=π at π, 1.57x=π  is a line parallel   to y-axis Consider, a vertical strip of length =y  and width =dx  in the first quadrantArea of the approximating rectangle =y dx   The approximating rectangle moves from  x=0 to x=π Area of the shaded region =0πy dx A=0πy dx A=0ππ 2+2 sin2x dxA=0ππ 2+21-cos 2x2 dxA=π 20πdx +0π1-cos 2x dxA=π 2x0π +x-sin 2x20πA=π 2π +  π -sin 2π 2-0A=π π 2+1A=π π+2 2A=π2 π+2  sq. unitsArea  of curve bound by x=0 and x=π is π2 π+2  sq. units

Page No 21.15:

Question 22:

Draw a rough sketch of the curve y=xπ+2 sin2 x and find the area between the x-axis, the curve and the ordinates x = 0 and x = π.

Answer:



The table for  different values of x and y is
 

x 0 π6 π2 5π6 π
sinx 0 12 1 12 0
y=xπ+2sin2 x 0 23 52 43 1


y=xπ+2 sin2 x   , is an arc cutting y-axis  at O(0, 0) and cutting x=π at π, 1 Consider a vertical strip of   length = y   and width = dx in the first quadrant Area of approximating rectangle =  y  dxThe approximating rectangle moves from  x=0  to x=πArea of the shaded area =0πy dxA=0πy dx                                                            As, y>0   y =yA=0πxπ+2 sin2 x dxA=1π0πx dx  +2 0πsin2 x dxA=1πx220π+2x2-12sin x cos x 0πA=π22π+22π-12sin π cos π  -0 A=π2+πA=3π2 sq. units Area of the curve enclosed between x=0 and x=π   is  3π2 sq. units 



Page No 21.16:

Question 23:

Find the area bounded by the curve y = cos x, x-axis and the ordinates x = 0 and x = 2π.

Answer:




 The shaded region is the required area bound  by the curve  y=cos x  , x axis   and x=0  , x=2πConsider  a vertical strip of length= y    and   width= dx  in the first quadrant Area of the approximating rectangle = y dxThe approximating rectangle moves from x=0 to x=2πNow ,0xπ2and 3π2x2π ,   y>0y =yπ2x3π2, y<0y =-yArea of the shaded region =02π y dxA=0π2 y dx +π23π2 y dx +3π22π y dxA=0π2 y dx +π23π2-y  dx +3π22π y dxA=0π2 cos x dx +π23π2-cos x dx +3π22πcos x dxA=sin x0π2+-sin xπ23π2+-sin x3π22πA=1+1+1+0--1A=4 sq. units Area bound  by the curve y=cos x, x-axis and x=0, x=2π=4 sq. units 

Page No 21.16:

Question 24:

Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x = π3 are in the ratio 2 : 3.

Answer:


  
The shaded area A1 is the required area bound by the curve y=sin x and x=0, x=π3The shaded area A2 is the required area bound by the curve y=sin 2x and x=0, x=π3 A1 =0π3y dxA1=0π3y dx               As, y>0 , y=yA1=0π3sin x  dxA1=-cos x0π3  A1=-cos π3+cos 0A1=-12+1=12                                          ... 1And, A2=0π3y dxA2=0π3y dx                    As, y>0 , y=yA2=0π3sin 2x  dxA2=0π32 sin x cos x dxA2=2-14cos 2x0π3A2=12-cos 2π3+cos 0A2=121+12 =12×32                                           ... 2From 1 and 2A1A2=12 12×32=23 Thus, the area of curve y=sin x and y=sin 2x for x=0 and x=π3  are in the ratio 2  :  3 

Page No 21.16:

Question 25:

Compare the areas under the curves y = cos2 x and y = sin2 x between x = 0 and x = π.

Answer:



 

Consider the value of  y  for different values of   x
 

  x 0 π4 π3 π2 2π3 5π6 π
y=cos2 x 1 0.5 0.25 0 0.25 0.75 1
y =sin2x 0 0.5 0.75 1 0.75 0.25 0


Let A1 be the area of curve y=cos2x  between x=0   and   x=π
Let A2 be the area of curve y=sin2 x between x=0  and x=π

Consider, a vertical strip of   length  =y   and width =dx in the shaded region of both the curves

The area of approximating rectangle =y dx

The approximating rectangle moves from  x=0 to x=πA1=0πy dxA1=0πy dx                                 0xπ ,  y>0 y=yA1=0πcos2 x dxA1=0π1+cos 2x dx                           cos2 x=1+cos 2x A1=12x+sin 2x20πA1=12π+sin 2π2-0A1=π2 Sq. unitsAlso,A2=0πy dxA2=0πy dx                                 0xπ ,  y>0 y=yA2=0πsin2 x dxA2=x2-12sin 2x20πA2=π2-12sin 2π2A2=π2  sq. unitsArea of curves  y=cos2 x  and area of curve  y=sin2 x  are both equal to  π2sq. units

Page No 21.16:

Question 26:

Find the area bounded by the ellipse x2a2+y2b2=1 and the ordinates x = ae and x = 0, where b2 = a2 (1 − e2) and e < 1.

Answer:



x2a2+y2b2=1  represents a parabola , symmetrical about both the axis .It cuts x axis at Aa, 0 and A'-a, 0It cuts y axis at B0, b and B'0, -bx=ae is a  line parallel to y axis Consider a vertical strip of length =y  and width=dx  ,in the first quadrantArea of approximating rectangle in first quadrant  = y dxApproximating rectangle moves from  x=0 to  x=aeArea of the shaded region =2 area in the first quadrant A=20aey dxA=20aey dx                                     As, y>0 , y=yA=20aebaa2-x2 dx                     x2a2+y2b2=1 y=baa2-x2 A=2ba0aea2-x2 dx A=2ba12xa2-x2  +12a2sin-1xa0aeA=2ba12aea2-a2e2  +12a2sin-1aea-0A=baa2e1-e2  +12a2sin-1eA=baa2e1-e2  +12sin-1eA=abe1-e2  +12sin-1e   sq. unitsRequired area  of the ellipse  bound by x=0 and x=ae  is= abe1-e2  +12sin-1e   sq. units

Page No 21.16:

Question 27:

Find the area of the minor segment of the circle x2+y2=a2 cut off by the line x=a2.

Answer:


The equation of the circle is x2+y2=a2.

Centre of the circle = (0, 0) and radius = a.

The line x=a2 is parallel to y-axis and intersects the x-axis at a2,0.


Required area = Area of the shaded region

                      = 2 × Area of the region ABDA

                      =2×a2aycircledx=2a2aa2-x2dx=2x2a2-x2+a22sin-1xaa2a=20+a22sin-11-a4×a2-a24+a22sin-112
                      =2a22×π2-a4×3a2-a22×π6=a2π2-3a24-a2π6=6a2π-33a2-2a2π12=a2124π-33 square units

Page No 21.16:

Question 28:

Find the area of the region bounded by the curve x=at2,y=2at between the ordinates corresponding t = 1 and t = 2.         [NCERT EXEMPLAR]

Answer:


The curve x=at2,y=2at represents the parametric equation of the parabola.

Eliminating the parameter t, we get

y2=4ax

This represents the Cartesian equation of the parabola opening towards the positive x-axis with focus at (a, 0).



When t = 1, x = a

When t = 2, x = 4a

∴ Required area = Area of the shaded region

                           = 2 × Area of the region ABCFA

                          =2a4ayparaboladx=2a4a4axdx=2×2a×x3232a4a=8a34a32-a32=8a38aa-aa=8a3×7aa=563a2 square units

Page No 21.16:

Question 29:

Find the area enclosed by the curve x = 3cost, y = 2sint.                    [NCERT EXEMPLAR]

Answer:


The given curve x = 3cost, y = 2sint represents the parametric equation of the ellipse.

Eliminating the parameter t, we get

x29+y24=cos2t+sin2t=1

This represents the Cartesian equation of the ellipse with centre (0, 0). The coordinates of the vertices are ±3,0 and 0,±2.



∴ Required area = Area of the shaded region

                          = 4 × Area of the region OABO

                         =4×03yEllipsedx=40341-x29dx=4×23039-x2dx=83x29-x2+92sin-1x303=830+92sin-11-0+0=83×92×π2=6π square units



Page No 21.24:

Question 1:

Find the area of the region in the first quadrant bounded by the parabola y = 4x2 and the lines x = 0, y = 1 and y = 4.

Answer:




y=4x2 represents a parabola , openeing upwards, symmetrical about +ve y-axis  and having vertex at O(0, 0)y=1 is a line parallel to x-axis , cutting parabola at -12 , 1 and 12, 1y=4    is a line parallel to x axis , cutting parabola at -1, 1 and 1, 1x=0 is the y-axis Consider a horizontal strip of  length= x and width=dy in the first quadrantArea of approximating rectangle =x dyApproximating rectangle moves from y=1 to y=4 Area of the curve in the  first quadrant enclosed by y=1 and y=4 is the required area of the shaded region Area of the shaded region =04x dyA=14x dy                As, x>0, x =xA=14y4 dy A=   1214y dy  A=12y323214A=12×23432-132A=138-1A=73 sq. unitsThe area  enclosed by parabola in the first quadrant  and y=1, y=4 is 73  sq. units

Page No 21.24:

Question 2:

Find the area of the region bounded by x2 = 16y, y = 1, y = 4 and the y-axis in the first quadrant.

Answer:



x2=16 y is a parabola, with vertex at O0, 0  and symmetrical about +ve y-axis y=1  is line parallel to x-axis cutting the parabola at -4, 1 and 4, 1 y=4  is line parallel to x axis cutting the parabola at -8, 1 and 8, 1 Consider a horizontal strip of length= x   and width =dy Area of approximating rectangle = x  dy The approximating rectangle moves from  y=1  to  y=4 Area of the curve in the first quadrant  enclosed by y=1 and y=4  is the shaded areaArea of the shaded region =14x  dyA=14x  dy                     As,  x>0,   x=xA=1416 y  dyA=414y  dyA=4y323214A=83432-132A=83×7=563 sq. unitsArea enclosed by parabola  in the first quadrant and y=1 and y=4  is 563 sq. units

Page No 21.24:

Question 3:

Find the area of the region bounded by x2 = 4ay and its latusrectum.

Answer:



x2=4ay represents a parabola with vertex O(0, 0) opening upwards and symmetrical about y-axis. F(0, a ) is the focus of the parabola and y=a its latus rectumConsider a horizontal strip of length =x and width dy in the first quadrant Area of approximating rectangle=x dyThe approximating rectangle moves from y=0 to y=a Area OAB =0ax dyArea OAA'O=2×Area OAB A=20ax dy =0ax dy               As, x>0 x=xA=20a4ay  dyA=2×2a0ay dyA=4a y32320aA=4a 23a32-0A=83a2 sq. units 

Page No 21.24:

Question 4:

Find the area of the region bounded by x2 + 16y = 0 and its latusrectum.

Answer:



x2+16 y=0 x2 =-16 yComparing it with equation of parabola x2=4aya=-4Thus, x2+16 y=0 represents a parabola, opening downwards, with vertex at O(0, 0) and -ve y-axis being its axis of symmetry Focus of the parabola is F(0, -4)y=-4 is the latus rectum of the parabolaThe latus rectum cuts the parabola at B(8, -4) and B'(-8, -4)x=8 cuts the x-axis at A(8, 0)Area of the curve bound by latus rectum= Shaded area BOB'B= 2 Area OBF                                                           ...1Consider a vertical strip of length= y and width=dx in shaded area OAB such that point P(x, y ) lies on the parabolaThe area of the approximating rectangle = y dxBut the approximating rectangle moves from x=0 to x=8Area of the shaded region OAB=08y dx A=08-x2 16 dx                     x2=-16 yy=-x2 16  Area of the shaded region OAB=08 x2 16 dx =116×13x308=8×8×816×3=323 sq. units                                       ...2So, area of rectangle OABF = OA×AB =8×4 =32 sq. units                                                                                               ...3 From 2   and 3Area of OBF=Area of rectangle OABF -Area of the shaded region OAB=32  -323=643  sq. units Shaded area BOB'B= 2 Area OBF=2×643=1283  sq. units Thus, area of the curve x2+16 y =0  bound by its latus rectum =1283  sq. units

Page No 21.24:

Question 5:

Find the area of the region bounded by the curve ay2=x3, the y-axis and the lines y = a and y = 2a.

Answer:


The equation of the given curve is ay2=x3.

The given curve passes through the origin. This curve is symmetrical about the x-axis.

The graph of the given curve is shown below.



The lines y = a and y = 2a are parallel to the x-axis and intersects the y-axis at (0, a) and (0, 2a), respectively.

∴ Required area = Area of the shaded region

                          =a2axcurvedy=a2aay213dy=a13a2ay23dy=a13×y5353a2a=35a132a53-a53=35253a2-a2=35253-1a2 square units



Page No 21.51:

Question 1:

Calculate the area of the region bounded by the parabolas y2 = x and x2 = y.

Answer:



y2=x is a parabola, opening sideways, with vertex at O(0, 0) and +ve x-axis as axis of symmetry x2 =y is a parabola, opening upwards, with vertex at O(0, 0) and +ve y-axis as axis of symmetry Soving the above two equations,x2=y4=y y4-y =0 y=0 or y=1 . So, x=0 or x=1O0, 0 and A(1, 1) are points of intersection of two curvesConsider a vertical strip of length = y2-y1  and width= dx Area of approximating rectangle =y2-y1 dx Approximating rectangle  moves from x=0 to x=1Area of the shaded region =01y2-y1 dx A=01 y2-y1 dx                     As,  y2-y1>0 y2-y1=y1A=01 x  -x2 dx A=x3232-x3301A=23×132-133   -0A=23-13A=13 sq. units Thus, area enclosed by the curves =13 sq. units

Page No 21.51:

Question 2:

Find the area of the region common to the parabolas 4y2 = 9x and 3x2 = 16y.

Answer:




3x2=16 y     ... 1 is a parabola with vertex at (0, 0) opening upwards and symmetrical about +ve y-axis 4y2=9x        ... 2 is a parabola with vertex at (0, 0) opening sideways and symmetrical about +ve x-axis Solving the equations 1 and 2, we get the points of intersection of the two parabolas O(0, 0) and A(4, 3 )   Consider a vertical strip of length= y2-y1 and width = dx such that P(x, y1) lies on 1 and Q(x, y2) lies on 2area of approximating rectangle =y2-y1  dx Approximating rectangle  moves from x=0  to x=4Area  of the shaded region =04y2-y1 dx =04y2-y1 dx                      As, y2-y1 =y2-y1 for y2-y1>0 A=0494x   -316x2 dx A=3204x dx  -31604x2 dx A=32x323204-316x3304A=432-116×43A=8-4=4 sq. units Area bound by the two curves = 4  sq. units  

Page No 21.51:

Question 3:

Find the area of the region bounded by y = x and y = x.

Answer:



y=x             ... 1  is a parabola opening side ways, with vertex at O(0, 0) and +ve x-axis as axis of symmetry x=y                  ...2 is a straight line passsing through O(0, 0) and at angle 45o with the x-axisSolving  1 and 2  y2=x=y y2=y y(y-1)=0 y=0 or y=1 and x=0 or x=1Thus, the line intersects the parabola at O(0, 0 ) and A(1, 1)Consider a approximating rectangle of length =y2-y1  and width= dx Area of approximating rectangle= y2 -y1  dxApproximating rectangle  moves from x=0 to x=1  Area of the shaded region=01 y2 -y1  dx =01 y2 -y1 dx              As, y2 >y1, y2 -y1=y2-y1 A=01 x  -x dx  A=x3232-x2201A=13232-122-0A=23-12=16  sq. units Area bound by the parabola and straight line = 16 sq. units

Page No 21.51:

Question 4:

Find the area bounded by the curve y = 4 − x2 and the lines y = 0, y = 3.

Answer:



y=4-x2 is a parabola, with vertex (0, 4), opening downwars and having axis of symmetry as -ve y-axis y=0   is the x-axis, cutting the parabola at A(2, 0) and A'(-2, 0)y=3 is a line parallel to x-axis, cutting the parabola at B(1, 3 ) and B'(-1, 3) and y-axis at C(0, 3)  Required area is the shaded area ABB'A =2 area ABCOConsider a horizontal strip of length= x2-x1 and width=dy in the shaded region Area of approximating rectangle= x2-x1 dyThe approximating rectangle moves from y=0 to y=3 Area of shaded region =  230x2-x1 dy A=203x2-x1 dy                       As,  x 2-x1 =x2-x1 , x2>x1 A=2034-y  -0 dyA=-24-y3232 03A=-24-y3232 03A=2 ×23432-132A=43×7  A=283  sq. unitsArea bounded by the two parabolas =283 sq. units

Page No 21.51:

Question 5:

Find the area of the region x, y:x2a2+y2b21xa+yb.

Answer:




Let R=x, y  :  x2 a2+y2b2 1xa+ybR1=x, y  :  x2 a2+y2b2 1and R2 =x, y  : 1xa+ybThen, R =R1R2Consider   x2 a2+y2b2=1. This represents  an ellipse, symmetrical about both axis and cutting  x-axis at  A(a, 0) and A'(-a, 0) and y-axis at B(0, b), B'(0, -b) R1= x2 a2+y2b2 1   represents the area inside the ellipse xa+yb=1 = represents a straight line cutting x-axis at  A(a, 0) and y-axis at B(0, b)R2=xa+yb1represents the area  above the straight line R =R1R2 represents the smaller shaded area bounded by the line and the ellipse In the shaded region, consider a vertical strip with length=y2-y1  and width =dx, such that P(x, y2) lies on ellipse and Q(x, y1) lies on the straight line Area of approximating rectangle=y2 -y1  dx The  approximating rectangle moves from  x=0 to x=aArea of the shaded  region =0ay2 -y1  dx =0ay2 -y1 dx                      As,  y2>y1, y2 -y1= y2-y1 A= 0abaa2-x2  -baa-x dxA=0abaa2-x2dx  - 0abaa-x  dxA=bax2a2-x2+12a2 sin-1xa0 a    -baax-x22A=ba0+12a2 sin-11-a2-a22A=ba12a2×π2-a22A=ab2π2-1A=ab4π-2 sq. units 

Page No 21.51:

Question 6:

Using integration, find the area of the region bounded by the triangle whose vertices are (2, 1), (3, 4) and (5, 2).

Answer:



Consider the points A(2, 1), B(3, 4) and C(5, 2) We need to find area of shaded triangle ABCEquation of AB is y-1=4-33-2 x-23x-y-5=0                                                         ... 1Equation of BC isy-4=2-45-3 x-3x=y-7=0                                                           ... 2Equation of CA is y-2=2-15-2 x-5x-3y+2=0                                                           ... 3Area of  ΔABC =Area of  ΔABD+Area of ΔDBC In ΔABD,  Consider point P(x, y2) on AB and Q(x, y1) on AD Thus, the area of approximating rectangle with length=y2-y1 and width=dx is y2-y1 dxThe approximating rectangle moves from x=2 to x=3 Area of ΔABD =23y2-y1 dx =23y2-y1  dx A=233x-5-x+13 dxA=239x-15-x-13 dx A=238x-163 dxA=138x22-16x23A=134×32-16×3-4×22+16×2A=1368-64A=43  sq. units Similarly, for S(x, y4) on AB and R(x, y3)  on DC Area of approximating rectangle of length  y4-y3 and width dx= y4-y3 dxApproximating rectangle  moves from x=3 to x=5Area BDC=35y4-y3  dxA=357-x -x+13 dxA=133520-4x dxA=1320 x-4x2235A=13100-50-60-18A=1350-42=83 sq. units Area of  ΔABC =Area of  ΔABD +Area of  ΔDBC=43+83=123 =4 sq. units

Page No 21.51:

Question 7:

Using integration, find the area of the region bounded by the triangle ABC whose vertices A, B, C are (−1, 1), (0, 5) and (3, 2) respectively.

Answer:




Equation of line AB isy-1=5-10+1x--1y=4x+5  Area under the line AB = area ABDO=-104x+5 dx=4x22+5x-10=0-2-5Area ABDO=3 sq. units                                             ... 1Equation of line BC is  y-5=2-53-0x-0y=-x+5 Area under line BC  =Area OBCP=03-x+5 dx=-x22+5x03=-92+15 -0Area OBCP=212  sq.  units                                     ... 2Equation of line CA isy-2=2-13--1x-3 4y =x+5Area under line AC  =Area ACPAA=-13x+54dxA=14x22+5x-13A=14322+5×3--122+5-1A=1492+15-12+5Area ACPA=244=6 sq. units                                    ... 3From 1, 2 and 3Area Δ ABC =Area ABDO+Area OBCP-Area ACPAA=3+212-6A=212-3 =21-62=152  sq. units Area Δ ABC=152  sq. units 

Page No 21.51:

Question 8:

Using integration, find the area of the triangular region, the equations of whose sides are y = 2x + 1, y = 3x + 1 and x = 4.

Answer:


Solving the given equations The point of intersection of the three lines are A(0, 1), B(4, 13) and C(4, 9). We need to find the area of ABCArea under line AB=area OABCLArea OABCL=043x+1 dx                    Equation of BC is y=3x+1 and x moves from A, x=0 to B, x=4 =3x22+x04 =3422+4=24+4=28 sq. unitsArea under line BC =Area OACLArea OACL=042x+1dx                       Equation of BC is y=2x+1 and x moves from A, x=0 to C, x=4 =2x22+x04=16+4=20 sq. unitsArea  ΔABC= Area OABCL-Area OACLArea  ΔABC=28-20 =8 sq. units Area of triangle formed by the three given lines=8 sq. units 

Page No 21.51:

Question 9:

Find the area of the region {(x, y) : y2 ≤ 8x, x2 + y2 ≤ 9}.

Answer:



Let R=x, y: y28x, x2+y29R1=x, y: y28xR2=x, y: x2+y29Thus, R=R1R2 Now, y2=8x  represents a parabola with vertex O(0, 0) and symmetrical about x-axis Thus, R1 such that y28x is the area inside the parabola Also, x2+y2=9 represents with circle with centre O(0, 0) and radius 3 units. The circle cuts the x axis at C(3, 0) and  C'(-3, 0 ) and Y-axis at B(0, 3) and B'(0, -3 )Thus, R2 such that x2+y29 is the area inside the circleR =R1R2 =Area OACA'O=2 shaded area OACO                       ... 1The point of intersection between the two curves  is obtained by solving the two equationsy2=8x  and x2+y2=9 x2+8x=9 x2+8x -9 =0x+9x-1=0x=-9  or  x=1Since, parabola is symmetric about +ve x-axis,  x=1  is the correct solution y2=8 y=±22Thus, A1, 22 and A' 1, -22  are the two points of intersectionArea OACO =area OADO +area DACD                       ... 2Area OADO  =018x dx                Area bound by curve y2=8x between x=0 and x=1=22x323201Area OADO=423                                                       ... 3Area DACD = area bound by x2+y2=9  between x=1 to x=3A=139-x2 dx  =12x9-x2+129 sin-1x313=0+92 sin-133 -129-12 -92 sin-113=92 sin-11 -128 -92 sin-113=92 π2-1222 -92 sin-113Area DACD=9 π4-2 -92 sin-113                            ... 4From 1, 2, 3 and 4R=Area OACA'O =2423+9 π4-2 -92 sin-113  =2423-2 +9 π4-92 sin-113  Area OACA'O =223 +9π4-92 sin-113 sq. units 

Page No 21.51:

Question 10:

Find the area of the region common to the circle x2 + y2 = 16 and the parabola y2 = 6x.

Answer:




 

Points of intersection of the parabola and the circle is obtained by solving the simultaneous equations

x2+y2=16 and y2=6xx2+6x=16 x2+6x-16 =0x+8x-2=0x=2  or  x=-8 , which is not the possible solution.  When x=2, y=±6×2 =±12=±23B2 ,23 and B'2 ,-23  are points of intersection of the parabola and circle. Now,Required area= areaOBAB'O                               =2areaOBAO                           =2areaOBDO+areaDBAD                           =2×026xdx +2416-x2 dx                           =2×6x323202+12x16-x2 +12×16 sin-1x424                           =2× 6×23×232-0 +12416-42 +12×16 sin-144-12×216-22 -12×16 sin-124                           =2 ×  6×23×22+0+8 sin-11-12 -8 sin-112                           =2×833+8×π2-23-8π6                           =2 83-633+8π2-π6                           =2233+82π6                           =433+16π3

Page No 21.51:

Question 11:

Find the area of the region between the circles x2 + y2 = 4 and (x − 2)2 + y2 = 4.

Answer:



x2+ y2 = 4   .......(1)       represents a circle with centre O(0,0) and radius  2
x-22+y2 = 4 ......(2)   represents a circle with centre A(2 ,0) and radius 2
 
Points of intersection  of two circles is given by solving the equations

x-22=x2x2-4x+4 =x2x=1 y2=3 y =±3Now, B1,3 and B'1,-3  are two points of intersection of the two circlesWe need to find shaded area= 2×areaOBAO     ...1AreaOBAO=areaOBPO+areaPBAP =01y1dx+12y2dx                                                   y1>0 y1=y1 and y2>0 y2=y2=01y1 dx+12y2dx=014-x-22 dx+124-x2 dx=12x-24-x-22 +12×4×sin-1x-2201+12x4-x2 +12×4×sin-1x212=-32+2sin-1-12 -0+2sin-1-1+0-123+2sin-11-sin-112=-32+2sin-1-12 -0-2sin-1-1+0-123+2sin-11-2sin-112=-3-4sin-112 +4sin-11=-3-4×π6 +4×π2=-3-2π3 +2π=4π3-3Now, From equation 1Shaded area= 2×areaOBAO=24π3-3=8π3-23  sq units 

Page No 21.51:

Question 12:

Find the area of the region included between the parabola y2 = x and the line x + y = 2.

Answer:



We have, y2=x and x+y=2
To find the intersecting points of the curves ,we solve both the equations.
 
y2+y-2=0y+2y-1=0y=-2 or y=1x=4 or 1Consider a horizantal strip of length x2-x1  and width dy where Px2,y lies on straight line and Qx1,y lies on the parabola.Area of approximating rectangle =x2-x1 dy , and it moves from y=-2 to y=1Required area = areaOADO =-21x2-x1 dy=-21x2-x1 dy             x2-x1=x2-x1 as x2>x1=-212-y-y2 dy=2y-y22-y33-21=2-12-13--4-2+83=2-12-13+6-83=92 sq units Area  enclosed by the line and given parabola =92 sq units             
                             
                                          
                                    

Page No 21.51:

Question 13:

Draw a rough sketch of the region {(x, y) : y2 ≤ 3x, 3x2 + 3y2 ≤ 16} and find the area enclosed by the region using method of integration.

Answer:



The  given region is the intersection of
          y23x   and  3x2+3y216
Clearly ,y2 = 3x is a parabola with vertex at (0, 0) axis is along the x-axis opening in the positive direction.
Also 3x2 + 3y2 = 16 is a circle with centre at origin and has a radius 163.
Corresponding equations of the given inequations are
y2=3x               .....1and3x2+3y2=16     .....2
 Substituting the value of y2 from (1) into (2)
  3x2+9x=16
3x2+9x-16=0
x=-9±81+1926
x=-9±2736
By figure we see that the value of x will be non-negative.
x=-9+2736
Now assume that x-coordinate of the intersecting point, a=-9+2736
The  Required area A  = 2(Area of OACO + Area of CABC)
Approximating the area of OACO  the length =y1  width = dx
    Area of   OACO  =0ay1 dx  
                            =0ay1 dx
                            =0a3xdx         y21=3x   y1=3x
                            =23 x323a0
                            =23a323 
Therefore, Area of  OACO =23a323
Similarly approximating the are of CABC the length =y2   and the width = dx
              Area of  CABC  =a43y2 dx     
                                     =a43y2 dx
                                     =a43163-x2 dx        3x2+3y22=16 y2=163-x2
                                     =x2163-x2 +83sin-1x34a43
                                    =423163-163+83sin-14343-a2163-a2-83sin-1a34
                        
                                   =-a2163-a2+83sin-11-83sin-1a34
 Area of  CABC =-a2163-a2+4π3-83sin-1a34
Thus the required area A = 2(Area of OACO + Area of CABC)
                                    =223a323-a2163-a2+4π3-83sin-1a34
                                   =4a323-a163-a2+8π3-163sin-1a34 
 Hence, the reguired area is,4a323-a163-a2+8π3-163sin-1a34 square units , where a=-9+2736
              
                             

Page No 21.51:

Question 14:

Draw a rough sketch of the region {(x, y) : y2 ≤ 5x, 5x2 + 5y2 ≤ 36} and find the area enclosed by the region using method of integration.

Answer:




The given region is intersection of
y25x  and  5x2+5y236
Clearly, y25x is a parabola with vertex at origin and the axis is along the x-axis opening in the positive direction.Also 5x2+5y236 is a circle with centre at the origin and has a radius 365 or  65.
Corresponding equations of given inequations are
         y2=5x                           .....1  5x2+5y2=36                .....2
Substituting the value of y2 from (1) into (2), we get

5x2+25x=36
5x2+25x-36=0
x=-25±625+72010
 x=-25±134510
From the figure we see that x-coordinate of intersecting point can not be negative.
x=-25+134510
Now assume that x-coordinate of intersecting point,  a=-25+134510
The  Required area,
A  = 2(Area of OACO + Area of CABC)
Approximating the area of OACO  the length =y1 and a width = dx=|y1=dx
Area of OACO=0ay1dx
               =0ay1 dx
               =0a5xdx         y12=5x   y1=5x
               = 52x323a0
Therefore, Area of OACO  =25a323
Similarly approximating the area of CABC the length =y2 and the width = dx
Area of CABC=a65y2dx 
               =a65y2 dx
               =a65365-x2dx
               =x2365-x2 +185sin-1x56a65
               =625365-365+185sin-11-a2365-a2-185sin-1a56
              =0+185sin-11-a2365-a2-185sin-1a56
              =185×π2-a2365-a2-185sin-1a56
Area of CABC  =9π5-a2365-a2-185sin-1a56
Thus the  Required area, A  = 2(Area of OACO + Area of CABC)
 A=2 25a323+ 9π5-a2365-a2-185sin-1a56
 =45a323+18π5-a365-a2-365sin-1a56 ,       Where,  a=-25+134510  .
  
   

Page No 21.51:

Question 15:

Draw a rough sketch and find the area of the region bounded by the two parabolas y2 = 4x and x2 = 4y by using methods of integration.

Answer:




To find the points of intersection between two parabola let us substitute   x=y24 in  x2=4y.

y242=4yy4-64y=0yy3-64=0y=0, 4
x=0, 4

Therefore, the points of intersection are A(4, 4) and C0, 0.

Therefore, the area of the required region ABCD =04y1dx-04y2dx where y1=2x and y2=x24

Required Area
 =042xdx-04x24dx=2×2x323-x31204=2×24323-4312-2×20323-0312

After simplifying we get,

=323-163=163 square units

Page No 21.51:

Question 16:

Find the area included between the parabolas y2 = 4ax and x2 = 4by.

Answer:


To find the point of intersection of the parabolas substitute y=x24b in y2=4ax we get

x416b2=4axx4-64ab2x=0xx3-64ab2=0x=0 and x=4ab23y=0 and y=4a2b3

Therefore, the required area ABCD04ab23y1-y2dx where y1=2ax and y2=x24b.

Required area =
 04ab23y1-y2dx=04ab232ax-x24bdx=4a3x32-x312b04ab23=4a34ab2332-4ab23312b-4a3032-0312a=16ab3square units



                      


    

Page No 21.51:

Question 17:

Prove that the area in the first quadrant enclosed by the x-axis, the line x = 3y and the circle x2 + y2 = 4 is π/3.

Answer:




x2+y2 =4  represents a circle  with centre O(0,0) and radius 2 , cutting x axis at A(2,0) and A'(-2,0)

x=3  y  represents a straight line passing through O(0,0)

Solving the two equations we get

x2+y2 =4 and x=3 y 3y2+y2=44y2 =4 y=±1x =±3B3 , 1 and B'-3 , -1  are points of intersection  of circle and straight lineShaded areaOBQAO= areaOBPO+area BAPB                                        =1303 x dx+324-x2 dx                                        =13x2203+12x4-x2+42sin-1x232                                        =32+0-32+2sin-1 1-sin-132                                        =32-32+2π2-π3                                        =π3 sq units     Area  bound by the circle and straight line above x axis =π3  sq units 

Page No 21.51:

Question 18:

Find the area of the region bounded by y=x,x=2y+3 in the first quadrant and x-axis.                 [NCERT EXEMPLAR]

Answer:


The curve y=x or y2=x represents the parabola opening towards the positive x-axis.

The curve x = 2y + 3 represents a line passing through (3, 0) and 0,-32.

Solving y2=x and x = 2y + 3, we get

y2=2y+3y2-2y-3=0y-3y+1=0y=3 or y=-1



∴ Required area = Area of the shaded region

                          =03xlinedy-03xparabolady=032y+3dy-03y2dy=2y+322×203-y3303=142×3+32-32-1333-0=1481-9-1327-0=18-9=9 square units

Page No 21.51:

Question 19:

Find the area common to the circle x2 + y2 = 16 a2 and the parabola y2 = 6 ax.


OR

Find the area of the region {(x, y) : y2 ≤ 6ax} and {(x, y) : x2 + y2 ≤ 16a2}.

Answer:





 

Points of intersection of the parabola and the circle is obtained by solving the simultaneous equations

x2+y2=16a2 and y2=6axx2+6ax=16a2 x2+6ax-16a2=0x+8ax-2a=0x=2a  or  x=-8a ,  x=-8a is not the possible solution.  When x=2a, y=±6a×2a =±12a=±23aB2a ,23a and B'2a ,-23a  are points of intersection of the parabola and circle. Now, Required area= areaOBAB'O                               =2×areaOBAO                           =2areaOBDO+areaDBAD                           =2×02a6axdx +2a4a16a2-x2 dx                           =2×6ax323202a+12x16a2-x2 +12×16a2 sin-1x4a2a4a                           =2× 6a×23×2a32-0 +12×4a16a2-4a2 +12×16a2 sin-14a4a-12×2a16a2-2a2 -12×16a2 sin-12a4a                           =2 ×  6a×23×2a2a+0+8a2sin-11-23a2 -8asin-112                           =2×8a233+8a2×π2-23a2-8a2π6                           =2 83-633a2+8π2-π6a2                           =2233a2+8a22π6                           =433a2+16π3a2                           =4a234π+3

Page No 21.51:

Question 20:

Find the area, lying above x-axis and included between the circle x2 + y2 = 8x and the parabola y2 = 4x.

Answer:



The given equations are x2+y2=8x        1 and y2=4x         2
Clearly the equation x2+y2=8x   is a circle with centre 4,0 and has a radius 4.Also y2=4x is a parabola with vertex at origin and the axis along the x-axis opening in the positive direction .
To find the intersecting points of the curves ,we solve both the equation.(2
 x2+4x=8x
x2-4x=0
xx-4=0
x=0 and x=4
When x=0, y=0
When x=4, y=±4

To approximate the area of the shaded region the length =y2-y1  and the width = dx
A=04y2-y1dx
  =04y2-y1dx   y2>y1   y2-y1= y2-y1
  =0416-x-42 - 4xdx    y2=16-x-42 and y1=2x
 
 =0416-x-42dx -044xdx
 =x-4216-x-42+ 162sin-1x-4440-4x32340
 =0+0-0-8sin-1-44-43×432
 =8π2-323=4π-323
Hence the required area is   4π-323  square units.


   

Page No 21.51:

Question 21:

Find the area enclosed by the parabolas y = 5x2 and y = 2x2 + 9.

Answer:



y=5x2    represents a parabola  with vertex at  O0,0    and opening upwards , symmetrical about +ve y axis y= 2x2+9 represents the wider parabola , with vertex at C0,9 To find point of intersection , solve the two equations 5x2=2x2+9 3x2=9x=±3y=15Thus A 3,15 and A'-3,15  are points of intersection of the two parabolas.Shaded area A'OA=2×areaOCAOConsider a vertical stip of length= y2- y1  and width=dx Area of approximating rectangle =y2- y1dx The  approximating rectangle moves from x=0 to x=3AreaOCAO =03y2- y1dx =03y2- y1dx               y2- y1 =y2- y1 as y2>y1=032x2+9-5x2dx        =039-3x2dx=9x-3x3303=93-33=63  sq unitsShaded area B'A'AB=2 area OCAO =2×63  =123  sq unitsThus area enclosed by two parabolas =123  sq units



Page No 21.52:

Question 22:

Prove that the area common to the two parabolas y = 2x2 and y = x2 + 4 is 323 sq. units.

Answer:


We have, two parabolas y = 2x2 and y = x2 + 4

To find point of intersection , solve the two equations 2x2=x2+4x2=4x=±2y=4Thus A (2,4) and A'(-2 ,4 )  are points of intersection of the two parabolasShaded area =2×areaOCAOConsider a vertical stip of length y2- y1 and width dx Area of approximating rectangle=y2- y1 dx The  approximating rectangle moves from x=0 to x=2AreaOCAO =02y2- y1dx =02y2- y1dx               y2- y1 =y2- y1 as y2>y1=02x2+4-2x2dx=024-x2dx=4x-x3302=8-82=163  sq unitsShaded area =2×areaOCAO =2×163  =323  sq units

Page No 21.52:

Question 23:

Using integration, find the area of the region bounded by the triangle whose vertices are (−1, 2), (1, 5) and (3, 4).            [CBSE 2014]

Answer:


Let ABC be the triangle with vertices A(−1, 2), B(1, 5) and C(3, 4).



Equation of AB is

y-5=2-5-1-1x-1y-5=32x-1y=32x+5-32=3x+72

Equation of BC is

y-4=5-41-3x-3y-4=-12x-3y=-12x+4+32=-x+112

Equation of CA is

y-2=4-23+1x+1y-2=12x+1y=12x+2+12=x+52

∴ Required area = Area of the shaded region

                           = Area of the region ABEFA + Area of the region BCDEB − Area of the region ACDFA

                          =-11yABdx+13yBCdx--13yCAdx=-113x+72dx+13-x+112dx--13x+52dx=12×3x+722×3-11+12×-x+1122×-113-12×x+522-13=112100-16-1464-100-1464-16=8412+364-484=7+9-12=4 square units

Page No 21.52:

Question 24:

Find the area of the region bounded by y=x and y = x.                [NCERT EXEMPLAR]

Answer:


The curve y=x or y2=x represents a parabola opening towards the positive x-axis.

The curve y = x represents a line passing through the origin.

Solving y2=x and y = x, we get

x2=xx2-x=0xx-1=0x=0 or x=1

Thus, the given curves intersect at O(0, 0) and A(1, 1).



∴ Required area = Area of the shaded region OAO

                          =01yparaboladx-01ylinedx=01xdx-01xdx=x323201-x2201=231-0-121-0=23-12=16 square units

Page No 21.52:

Question 25:

Find the area of the region in the first quadrant enclosed by x-axis, the line y = 3x and the circle x2 + y2 = 16.

Answer:





x2 + y2 =16  represents a circle with centre O(0,0) and cutting the x axis at A(4,0)

y=3 x represents straight passing through O(0,0)

Point of intersection is obtained by solving the two equations

x2+y2 =16 and y=3 x x2+3 x2 =164x2 =16 x=±2y =±23B2 ,23 and B'-2 ,-23  are points of intersection  of circle and straight lineShaded areaOBQAO= areaOBPO+area PBQAP                                        =023 x dx+2416-x2 dx                                        =3x2202+12x16-x2+162sin-1x424                                        =23 +8×π2-23-8×π6                                        =23 +4π-23-4π3                                        =8π3  sq units     Area  bound by the circle and straight line above x axis =23 +-23+8×2π6    =8π3  sq units 

Page No 21.52:

Question 26:

Find the area of the region bounded by the parabola y2 = 2x + 1 and the line xy − 1 = 0.

Answer:



 

We have,  y2 =2x+1 and x-y-1=0 

To find the intersecting points of the curves ,we solve both the equations.
 
y2=21+y+1y2-2-2y-1=0y2-2y-3=0y-3y+1=0y=3 or y=-1x=4 or 0Consider a horizantal strip of length x2-x1  and width dy where Px2,y lies on straight line and Qx1,y lies on the parabola.Area of approximating rectangle =x2-x1 dy , and it moves from y=-1 to y=3Required area = areaOADO =-13x2-x1 dy=-13x2-x1 dy             x2-x1=x2-x1 as x2>x1=-131+y-12y2-1dy=-131+y-12y2+12dy=-1332+y-12y2dy=32y+y22-16y3-13=92+92-276--32+12+16=92+56=163 sq units Area  enclosed by the line and given parabola =163 sq units             
                             
                                          
                                    

Page No 21.52:

Question 27:

Find the area of the region bounded by the curves y = x − 1 and (y − 1)2 = 4 (x + 1).

Answer:


We have, y = x − 1 and (y − 1)2 = 4 (x + 1)

x-1-12=4x+1x-22=4x+1x2+4-4x=4x+4x2+4-4x-4x-4=0x2-8x=0x=0 or x=8y=-1 or 7Consider a horizantal strip of length x2-x1  and width dy where Px2,y lies on straight line and Qx1,y lies on the parabola.Area of approximating rectangle =x2-x1 dy , and it moves from y=-1 to y=7Required area = areaOADO =-17x2-x1 dy=-17x2-x1 dy             x2-x1=x2-x1 as x2>x1=-171+y-14y-12-4dy=-171+y-14y-12+1dy=-172+y-14y-12dy=2y+y22-112y-13-17=14+492-112×6×6×6--2+12+112×2×2×2=14+492-18--2+12+23=412+56=643 sq units Area  enclosed by the line and given parabola =643 sq units

Page No 21.52:

Question 28:

Find the area enclosed by the curve y=-x2 and the straight line x + y + 2 = 0.                [NCERT EXEMPLAR]

Answer:


The curve y=-x2 represents a parabola opening towards the negative y-axis.

The straight line x + y + 2 = 0 passes through (−2, 0) and (0, −2).

Solving y=-x2 and x + y + 2 = 0, we get

x-x2+2=0x2-x-2=0x-2x+1=0x=2 or x=-1

Thus, the parabola y=-x2 and the straight line x + y + 2 = 0 intersect at A(−1, −1) and B(2, −4).



∴ Required area = Area of the shaded region OABO

                          =-12ylinedx--12yparaboladx=-12-x+2dx--12-x2dx=-x+222-12+x33-12=-1216-1+138--1=-152+3=-92=92 square units

Page No 21.52:

Question 29:

Find the area bounded by the parabola y = 2 − x2 and the straight line y + x = 0.

Answer:

The graph of the parabola y=2-x2 and the line x+y=0 can be given as:



To find the points of intersection between the parabola and the line let us substitute   x=-y in y=2-x2.

y=2-y2y2+y-2=0y-1y+2=0y=1, -2
x=-1, 2

Therefore, the points of intersection are A(-1, 1) and C2, -2.

The area of the required region ABCD =-12y1dx--12y2dx where y1=2-x2 and y2=-x

Required Area
 =-122-x2+xdx=2x-x33+x22-12=22-233+222-2-1--133+-122

After simplifying we get,

=92 square units

Page No 21.52:

Question 30:

Using the method of integration, find the area of the region bounded by the following lines:
3xy − 3 = 0, 2x + y − 12 = 0, x − 2y − 1 = 0.

Answer:



We have,

3x -y-3=0         12x+y-12=0       ...2x-2y -1=0         ...3


Solving 1 and 2, we get, 5x-15=0 x=3  y =6B(3,6) is point of intersection of 1 and 2Solving 1 and 3, we get, 5x=5x=1 y=0A1,0 is point of intersection of 1 and 3Solving 2 and 3, we get, 5x=25 x=5 y=2C5,2 is point of intersection of 2 and 3Now, Area ABC =area bound by 1 between x=1 and x=3 +area bound by 2 between x=3 and x=5 -area bound by 3 between x=1 and x=5                    =133x-3 dx+3512-2xdx-15x-12dx                   =3×x22-3x13+12x-2×x2235-12x22-x15                   =3x22-x13+12x-x235-12x22-x15                   =392-3-312-1+12-25-12-9-12252-1-12-1                   =6+8-4                   =10 sq units

Page No 21.52:

Question 31:

Sketch the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 1. Also, find the area of this region.

Answer:




We have, y=x2+2   and y=x
We see that parabola and the line y=x do not intersect
x =1    is a line parallel to y axis

Point of intersection between parabola and  x=1 is
Putting x=1 in y=x2+2, we get,y=1+2=3Point of intersection of two lines  is given by Putting x=1 in y=x, we get,y=1  Consider a vetical strip of length y2-y1  and width=dx such that Px,y2  lies on parabola and Qx ,y1 lies on y=xShaded area=01y2-y1 dx=01y2-y1 dx              y2-y1 y2-y1as y2>y1 ,=01x2 +2-x dx=01x2 +2-xdx=x3 3-x2 2+2x01=13-12+2=2-3+126=116sq units Thus, area enclosed by parabola and given two lines =116 sq units 

Page No 21.52:

Question 32:

Find the area bounded by the curves x = y2 and x = 3 − 2y2.

Answer:




 x = y2 is a parabola opening towards positive x-axis , having vertex at O (0,0) and symmetrical about x-axis
 x = 3 − 2y2 is a parabola opening negative x-axis, having vertex at A (3, 0) and symmetrical about x-axis, cutting y-axis at B and B'

Solving the two equations for the point of intersection of two parabolas

x =y2x=3-2y2y2 =3-2y2 3y2 =3y =±1y=1 ,x=1  and y=-1 x=1E1,1  and  F1,-1are  two points of intersection. The curve character changes at E and F .Draw EF parallel to y-axis.C(1, 0) is the point of intersection of EF ith x-axisSince both curves are symmetrical about x-axis ,Area of shaded region OEAFO = 2 Area OEAO=2Area OECO +area CEAC    .....1Area OECO =o1y1 dx       where Px,  y1 is a point on x=y2 =o1y1 dx                     as y1 >0=o1x dx=x323201 =23   sq units    .....2area CEAC =13y2 dx     where   Qx, y2 is a point on x=3-2y2 =13y2 dx        as y2 >0  =133-x2dx=12133-x dx=12-3-x323213=12×230+232=2×222×3=43  sq. units    .....3  From 1, 2 and 3, we get Therefore, area of Shaded region OEAFO =223 +43 =2×2 =4 sq units

Page No 21.52:

Question 33:

Using integration, find the area of the triangle ABC coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4).

Answer:




A(4, 1), B(6, 6) and C(8, 4) are three given points.
Equation of AB  is given byy-1 =6-16-4x-4y-1 =52x-4y=52x-9        .....1Equation of BC is given byy-6 =4-68-6 x-6y-6 =-x-6y=-x+12        .....2Equation of CA is given byy-4=1-44-8x-8y-4=34x-8y=34x-2        .....3
Required area of shaded region ABC
Shaded areaABC = areaABD+ areaDBC Consider a point Px, y2on AB  and Qx, y1  on ADfor a vertical strip of length =y2-y1  and  width= dx, area =y2-y1dx,The approximating rectangle moves from x=4 to x=6Hence  areaABD =46y2-y1dx=4652x-9-34x-2dx=4674x-7dx=74×x22-7x46=7836-16-76-4=78×20 -14=352-14=35-282 =72 sq units

Consider a point Sx, y4 on BC  and Rx, y3  on DC For a vertical strip of length =y4-y3  and  width= dx, area =y4-y3dx,The approximating rectangle moves from x=6 to x=8area BDC=68-x+12-34x-2 dx=68-74x +14dx=- 74×x22 +14x68=- 7x28+14x68=-7864-36 +148-6=-78×28 +28=28 8=7 2Thus required areaABC=72+72=7 sq. units

Page No 21.52:

Question 34:

Using integration find the area of the region: x, y : x-1y5-x2.

Answer:



x-1y5-x2x-1=5-x2x=2,-1A=-125-x2-x-1dx=-125-x2+-11x-1dx+121-xdx=x25-x2+52sin-1x5-12+x22-x-11+x-x2212=52sin-125+sin-115+12

Page No 21.52:

Question 35:

Find the area of the region bounded by y = | x − 1 | and y = 1.

Answer:



We have,y=x-1y=x-1       x11-x      x<1
y = x − 1 is a straight line originating from A(1, 0)  and making an angle 45o with the x-axis
y = 1 − x is a straight line originating from A(1, 0)  and making an angle 135o with the x-axis

y = x is a straight line parallel to x-axis and passing through B(0, 1)

The point of intersection of  two lines with y = 1 is obtained by solving the simultaneous equations

y=1and y=x-1 1=x-1x-2=0x=2C2, 1  is point of intersection of y=x-1 and y=1y=1  and y=1-x1=1-xx=0B0, 1  is point of intersection of y=1-x and y=1Since y= x-1  changes character at A(1, 0) ,Consider point P (1,1) on BC such that PA is perpendicular to x-axis.Required shaded area ABCA = area ABPA+area PCAP=011-1-xdx +121-x-1dx=01x dx +122-x dx=x2201+2x-x2212=12+4-2-2+12=12+12=1  sq. unit 

Page No 21.52:

Question 36:

Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32.

Answer:




We have, x2+y2 =32 and y=x

 The point of intersection of the circle and parabola is obtained by solving the two equations

x2+x2=322x2=32   x2=16    x=±4 y=±4  Thus C4, 4   and C'-4, -4  are points of intersection of  the circle and straight line.Required shaded areaOCAPO = areaOCPO+areaPCAP=04y1dx+432y2dx=04y1 dx+432y2 dx                   y1>0 y1=y1 and y2>0 y2=y2=04x dx+43232-x2 dx =x2204+12x32-x2+12×32  sin-1x32432= 8+8π -8-4π=4π sq units.

Page No 21.52:

Question 37:

Find the area of the circle x2 + y2 = 16 which is exterior to the parabola y2 = 6x.

Answer:




 

Points of intersection of the parabola and the circle is obtained by solving the simultaneous equations

x2+y2=16 and y2=6xx2+6x=16 x2+6x-16 =0x+8x-2=0x=2  or  x=-8 , which is not the possible solution.  When x=2, y=±6×2 =±12=±23B2 ,23 and B'2 ,-23  are points of intersection of the parabola and circle. Required area = AreaOB'C'A'CBO=  area of circle -areaOBAB'O    Area of circle with radius 4 =π×42=16π         Now, Area OBAB'O =2areaOBAO                           =2areaOBDO+areaDBAD                           =2×026xdx +2416-x2 dx                           =2×6x323202+12x16-x2 +12×16 sin-1xa24                           =2× 6×23×232-0 +12416-42 +12×16 sin-144-12×216-22 -12×16 sin-124                           =2 ×  6×23×22+0+8 sin-11-12 -8 sin-112                           =2×833+8×π2-23-8π6                           =2 83-633+8π2-π6                           =2233+82π6                           =433+16π3Shaded area =16π-433+16π3=48π-16π3- 433=32π3- 433=438π-3  sq units 

Page No 21.52:

Question 38:

Find the area of the region enclosed by the parabola x2 = y and the line y = x + 2.

Answer:






 The points of intersection C and D are obtained by solving the two equations

x2=x+2x2-x-2=0x-2x+1=0x=2  or x=-1y=22=4  or y=-12=1Thus C(2,4 and D-1,1   are the points of intesection of two curvesConsider a vertical steip of length y2-y1  and width dx where Px,y2 lies on straight line and Qx,y1 lies on the parabola.Area of approximating rectangle =y2-y1 dx , and it moves from x=-1 to x=2Required area = areaODBCO =-12y2-y1 dx=-12y2-y1 dx             y2-y1=y2-y1 as y2>y1=-12x+2-x2 dx=-12x+2-x2 dx=x22+2x-x33-12=42+4-83-12+2-13=92 sq units Area  enclosed by the line and given parabola =92 sq units

Page No 21.52:

Question 39:

Make a sketch of the region {(x, y) : 0 ≤ yx2 + 3; 0 ≤ y ≤ 2x + 3; 0 ≤ x ≤ 3} and find its area using integration.

Answer:




R=x,y:0yx2+3  ,  0y2x+3  , 0x3R1=x,y :0yx2+3  R2=x,y : ,  0y2x+3   R3=x,y : 0x3  R=R1R2R3

y = x2 + 3 is a upward opening  parabola with vertex A(0, 3).
Thus R1 is the region above x-axis and  below the parabola
y = 2x + 3 is a straight line passing through A(0, 3) and cuts y-axis on (−3/2, 0).
Hence R2 is the region above x-axis and below the line
x = 3  is a straight line parallel to y-axis, cutting x-axis at E(3, 0).
Hence R3 is the region above x-axis  and to the left of the line x = 3.
Point of intersection of the parabola  and  y = 2x + 3 is given by solving the two equations
y=x2+3y=2x+3x2+3=2x+3x2-2x=0xx-2=0x=0  or x=2y=3 or y=7A0, 3 and B2, 7  are points of intersectionAlso ,x=3 cuts the parabola at C3, 12  and x=3 cuts  y=2x+3   at D3, 9 We require thearea of shaded region. Total shaded area =02x2+3dx +232x+3 dx=x33+3x02+2x22+3x23=x33+3x02+x2+3x23=83+6+9+9-4-6=83+6+8=8+423=503 sq. units

Page No 21.52:

Question 40:

Find the area of the region bounded by the curve y = 1-x2, line y = x and the positive x-axis.

Answer:


y=1-x2 y2 =1-x2x2+y2 =1 
Hence, y=1-x2  represents the upper half of the circle x2 + y2 = 1 a circle with centre O(0, 0) and radius 1 unit.
y = x   represents equation of a straight line passing through O(0, 0)
Point of intersection is obtained by solving two equations

  y=xy=1-x2x=1-x2x2=1-x22x2=1 x=±12 y=±12D12,12  and D'-12,-12  are two points of intersection between the circle and the straight lineAnd D12, 12 is  the intersection point of y=1-x2 and y=x.Required area=Shaded area ODAEO =Area ODEO+ area EDAE                   .....1Now, area ODEO =012x dx=x22012 =12122=14 sq units                   .....2Area EDAE =1211-x2 dx=12x1-x2+×12×sin-1x1121=0+12sin-11 -12×12×1-122 -12sin-112=12×π2-14-12×π4                                       using, sin-11=π2 and sin-112=π4=π4-π8-14=π8-14    sq units                   .....3From 1, 2 and 3, we getAreaODAEO =14+π8-14   =π8 sq. units

Page No 21.52:

Question 41:

Find the area bounded by the lines y = 4x + 5, y = 5 − x and 4y = x + 5.

Answer:


We have,y=4x+5           .....1y=5-x            .....24y=x+5          .....3
All the three equations represent equations of straight lines
The points of intersection is obtained by solving  simultaneous equations
From 1 and 24x+5=5-x5x=0x=0 y=5 Thus  A0, 5 is the point of intersection of 1  and 2From 2  and 345-x=x+55x=15x=3y=2Thus B3, 2 is the point of intersection of 2 and 3From 1 and 344x+5=x+515x=-15x=-1y=1Thus  C-1, 1 is  the point of intersection of 1  and 3Area ABC =areaABP +area PAB=-104x+5-x+54 dx +035-x-x+54 dx=-10154x+154 dx +03154-54x dx=154x22+x-10 +543x-x2203=154-12+1 +549-92=158+54×92=158+458 =608=152  sq. units

Page No 21.52:

Question 42:

Find the area of the region enclosed between the two curves x2 + y2 = 9 and (x − 3)2 + y2 = 9.

Answer:



Let the two curves be named as y1 and y2 where
y1:x-32+y2=9       .....1y2:x2+y2=9       .....2
The curve x2 + y2 = 9 represents a circle with centre (0, 0) and the radius is 3.
The curve (x − 3)2 + y2 = 9 represents a circle with centre (3, 0)  and has a radius 3.
To find the intersection points of two curves equate them.
On solving (1) and (2) we get
x=32 and y=±332
Therefore, intersection points are 32, 332 and 32, -332 .
Now, the required area(OABO) =2[area(OACO) +area(CABC)]
Here,
 AreaOACO=032Y1dx
=0329-x-32dx
And
AreaCABC=323Y2dx
=3239-x2dx
Thus the required area is given by,
A = 2[area(OACO) +area(CABC)]
20329-x-32 dx + 3239-x2dx
=2x-329-x-32+92sin-1x-33032+2x29-x2+92sin-1x3332
=232-329-33-32 +92sin-132-33-0-329-0-32-92sin-10-33+2329-32+92sin-133-349-94-92sin-1323

=2-938-9π12+9π4+29π4-938-9π12 

=-1838-18π12+18π4+18π4-1838-18π12

=-3638-36π12+36π4 

=-932-3π+9π
=6π-932     
Hence the required area is 6π-932  square units.

Page No 21.52:

Question 43:

Find the area of the region {(x, y): x2 + y2 ≤ 4, x + y ≥ 2}.

Answer:



Let R=x, y: x2+y24  ,x+y2R1=x, y: x2+y24R2=x, y: x+y2R=R1R2
The region R1 represents interior of the circle x2 + y2 = 4 with centre (0, 0) and has a radius 2.
The region R2 lies above the line x + y =2
The line x + y =2 and circle x2 + y2 = 4 intersect each other at (2, 0) and (0, 2).
Here, the length of the shaded region is given by y2-y1 where y2 is y for the circle x2 + y2 = 4 and y1 is y for the line x + y = 2 ; y2 > y1 and  the width of the shaded portion is dx.
Therefore the area,
A=02y2-y1dx=024-x2-2-xdx=12x4-x2+42sin-1x202-2x-x2202=224-22+42sin-122-024-02-42sin-102-22-222-20+022=0+2sin-11-0-0-4-2-0+0=2sin-11-2=2×π2-2=π-2  

Page No 21.52:

Question 44:

Using integration, find the area of the following region: x, y :x29+y241x3+y2.

Answer:



Let R=x, y : x29+y241x3+y2R1=x,y : x29+y241 and R2=x,y : 1 x3+y2R=R1 R2x29+y24=1  represents an ellipse, with centre at O(0, 0) , cutting the coordinate axis at A3, 0, A'-3, 0, B0, 2 and B'0, -2Hence, R1 is area interior to the ellipse x3+y2=12x+3y=6   represents a straight line cutting the coordinate axis at A3, 0 and B0, 2Hence,R2 will be area above the lineA3, 0 and B0, 2 are points of intersection of ellipse and straight line.Area of shaded region,A=0341-x29-21-x3 dx=0336-4x29-6-2x3dx=130329-x2-6-2x dx=132×12x9-x2+12×9 sin-1x3-6x-2x2203=13x9-x2+9 sin-1x3-6x+x203=130+9sin-11-18+9-0      =139π2-9=3π2 -3 sq units



Page No 21.53:

Question 45:

Using integration find the area of the region bounded by the curves y=4-x2,  x2+y2-4x=0 and the x-axis.

Answer:

The given curves are y=4-x2 and x2+y2-4x=0.
y = 4-x2x2+y2=4          .....(1)
This represents a circle with centre O(0, 0) and radius = 2 units.
Also,
x2+y2-4x=0x-22+y2=4         .....(2)
This represents a circle with centre B(2, 0) and radius = 2 units.
Solving (1) and (2), we get
x-22=x2x2-4x+4 =x2x=1 y2=3y =±3
Thus, the given circles intersect at A1, 3 and C1, -3.


∴ Required area
= Area of the shaded region OABO
=014-x-22 dx+124-x2 dx=12x-24-x-22 +42sin-1x-2201+12x4-x2 +42sin-1x212=-32+2sin-1-12 -0+2sin-1-1+0-123+2sin-11-sin-112
=-32-2×π6+2×π2-32+2×π2-2×π6=-3+2π -2π3=4π3-3 square units

Page No 21.53:

Question 46:

Find the area enclosed by the curves y = | x − 1 | and y = −| x − 1 | + 1.

Answer:




The given curves are
                      y=x-1         .....1
                y=-x-1+1          .....2
Clearly y=x-1 is cutting the x-axis  at (1, 0) and the y-axis at (0, 1) respectively.
Also y=-x-1+1 is cutting both the axes at (0, 0) and x-axis at (2, 0).
We have,
y=x-1y=x-1       x11-x       x<1Andy=-x-1+1y=2-x       x1  x       x<1Solving both the equations for x<1y=1-x and y=x,We get x=12 and y=12And solving both the equations for x1y=x-1 and y=2-x,We get x=32 and y=12
Thus  the intersecting points are 12, 12 and 32, 12.
The required area A = ( Area of  ABFA  + Area of  BCFB)
Now approximating the area of ABFA  the length = y1   and width  = dx
 Area of  ABFA 
=121x-1-xdx=1212x-1dx=x2-x121=14
Similarly approximating the area of  BCFB the length =y2   and  width= dx
Area of  BCFB
=1322-x-x-1 dx=1323-2x dx=3x-x2132=14
Thus the required area A =( Area of ABFA  + Area of BCFB)
                                      =14+14=12
Hence the required area is 12 square units.

                    

Page No 21.53:

Question 47:

Find the area enclosed by the curves 3x2 + 5y = 32 and y = | x − 2 |.

Answer:



3x2+5y=32   represents a downward opening parabola, symmetrical about negative x-axis

 vertex of the parabola is D 0, 325 . It cuts the X axis at B323   ,0 and B'-323   ,0Also,y=x-2  y=x-2 ,  x22-x  , x<2  y=x-2 ,  x2     represents a line , cutting the x axis at  A2, 0 and the parabola at C 3, 1And y=2-x, x<2 represents a line, cutting the parabola at -2, 4Shaded area AEYCA =-2232-3x25-2-xdx  +2332-3x25-x-2dx     =-2332-3x25dx --222-xdx +23x-2dx=1532x-x3-23-2x-x22-22-x22-2x23=1532×3-27+64-8-4-2+4+2-92-6-2+4=1596-35+64-8-92-4=1255-4-92=25-4-92=21-92=42-92=332 sq units

Page No 21.53:

Question 48:

Find the area enclosed by the parabolas y = 4xx2 and y = x2x.

Answer:




We have, y=4x-x2  and y=x2-x
The points of intersection of two curves is obtained by solving the simultaneous equations

x2 -x=4x-x22x2 -5x =0 x=0 or x=52y=0 or y=154O0,0 and D 52  ,154 are points of intersection of two parabolas.In the shaded area CBDC , consider P(x,y2) on  y =4x-x2  and Q(x,y1) on y=x2 -xAreaOBDCO= areaOBCO +areaCBDC=01y dx +152y2 -y1 dx=01y dx +152y2 -y1 dx        y>0 y  =y and y2 -y1y2-y1 as y2>y1   =014x-x2dx +1524x-x2-x2 -xdx=4x22-x3301+1525x-2x2dx=2x2 -x3301+5x22-2x33152=2-13+52522-23523 -52+23=53+5231-23 -116=53+52313-116=10-116+12524 =12124 sq units          ...1AreaOCV'O =01y  dx =01-y dx        y<0 y  =-y=01-x2 -xdx=01x-x2  dx=x22-x3 301=12-13=16 sq units          ...2From 1  and 2Shaded area=areaOBDCOand areaOCV'O                 =12124+16                 =12524 sq units

Page No 21.53:

Question 49:

In what ratio does the x-axis divide the area of the region bounded by the parabolas y = 4xx2 and y = x2x?

Answer:




We have, y=4x-x2  and y=x2-x
The points of intersection of two curves is obtained by solving the simultaneous equations

x2 -x=4x-x22x2 -5x =0 x=0 or x=52y=0 or y=154O0,0 and D 52  ,154 are points of intersection of two parabolas.In the shaded area CBDC , consider P(x,y2) on  y =4x-x2  and Q(x,y1) on y=x2 -xWe need to find ratio of areaOBDCOand areaOCV'OAreaOBDCO= areaOBCO +areaCBDC=01y dx +152y2 -y1 dx=01y dx +152y2 -y1 dx        y>0 y  =y and y2 -y1y2-y1 as y2>y1   =014x-x2dx +1524x-x2-x2 -xdx=4x22-x3301+1525x-2x2dx=2x2 -x3301+5x22-2x33152=2-13+52522-23523 -52+23=53+5231-23 -116=53+52313-116=10-116+12524 =12124 sq units          ...1AreaOCV'O =01y  dx =01-y dx        y<0 y  =-y=01-x2 -xdx=01x-x2  dx=x22-x3 301=12-13=16 sq units          ...2From 1  and 2 Ratio=AreaOBDCOAreaOCV'O=1212416=1214=121: 4  

Page No 21.53:

Question 50:

Find the area of the figure bounded by the curves y = | x − 1 | and y = 3 −| x |.

Answer:


We have,y=x-1y=x-1      for x11-x      for x<1
y = x − 1 is a straight line passing through A(1, 0)
y = 1 − x is straight line passing through A(1, 0) and cutting y-axis at  B(0, 1)

y=3-xy=3-x                          for xo3--x  =3+x       for x<0

y = 3 − x is straight line passing through C(0, 3) and D(3, 0)
y = 3 + x is a straight line passing through C(0, 3) and  D'(−3, 0)
The point of intersection is obtained by solving the simultaneous equations
y=x-1and y=3-xWe getx-1=3-x2x-4 =0x=2y=2-1=1Thus P2, 1 is point of intersection of  y=x-1 and y=3-xPoint of intersection for y=1-xy=3+x1-x=3+x2x=-2x=-1y=1--1 =2 Thus Q-1, 2 is point of intersection of y=1-x and y=3+xSince the character of function changes at C0, 3 and A(1, 0) , draw AM perpendicular to x-axis Required area =Shaded areaQCPAQ= AreaQCB+AreaBCMAB+areaAMPA  .....1 AreaQCB =-103+x -1-xdx=-102+2x dx=2x+x2-10= 0--2+1=1   sq unit       .....2AreaBCMA =013-x-1-x dx=012 dx =2x01 =  2 sq unit       .....3AreaAMPA =123-x-x-1 dx=124-2x dx=4x-x212 =8-4 -4-1=1  sq unit         .....4From 1, 2, 3 and 4Shaded area =1+2+1 =4  sq units

Page No 21.53:

Question 51:

If the area bounded by the parabola y2=4ax and the line y = mx is a212 sq. units, then using integration, find the value of m.

Answer:


The parabola y2=4ax opens towards the positive x-axis and its focus is (a, 0).

The line y = mx passes through the origin (0, 0).

Solving y2=4ax and y = mx, we get

m2x2=4axm2x2-4ax=0xm2x-4a=0x=0 or x=4am2
So, the points of intersection of the given parabola and line are O(0, 0) and A4am2,4am.


∴ Area bounded by the given parabola and line

= Area of the shaded region

=04am2yparaboladx-04am2ylinedx=04am24axdx-04am2mxdx=2a×x323204am2-m×x2204am2=4a34am232-0-m24am22-0=4a3×8aam3-m2×16a2m4=32a23m3-8a2m3=8a23m3 square units

But,
Area bounded by the given parabola and line = a212 sq. units          (Given)
8a23m3=a212m3=32m=323

Thus, the value of m is 323.

Page No 21.53:

Question 52:

If the area enclosed by the parabolas y2 = 16ax and x2 = 16ay, a > 0 is 10243 square units, find the value of a.

Answer:


The parabola y2 = 16ax opens towards the positive x-axis and its focus is (4a, 0).

The parabola x2 = 16ay opens towards the positive y-axis and its focus is (0, 4a).

Solving y2 = 16ax and x2 = 16ay, we get

x216a2=16axx4=16a3xx4-16a3x=0xx3-16a3=0x=0 or x=16a

So, the points of intersection of the given parabolas are O(0, 0) and A(16a, 16a).



Area enclosed by the given parabolas

= Area of the shaded region

=016a16axdx-016ax216adx=4a×x3232016a-116a×x33016a=8a316a32-0-148a16a3-0=8a3×64aa-256a23=512a23-256a23=256a23 square units

But,

Area enclosed by the given parabolas = 10243 square units                   (Given)
256a23=10243a2=1024256=4a=2           a>0

Thus, the value of a is 2.



Page No 21.61:

Question 1:

Find the area of the region between the parabola x = 4yy2 and the line x = 2y − 3.

Answer:



To find the point of intersection of the parabola x = 4yy2 and the line x = 2y − 3
Let us substitute x = 2y − 3 in the equation of the parabola.
2y-3=4y-y2y2-2y-3=0y+1y-3y=-1, 3
Therefore, the points of intersection are D(−1, −5) and A(3, 3).
The area of the required region ABCDOA,
  A=-13x1-x2 dy    where, x1=4y-y2 and x2=2y-3=-13x1-x2 dy           x1>x2=-134y-y2-2y-3 dy=-134y-y2-2y+3 dy=-13-y2+2y+3dy=-y33+2y22+3y-13=-y33+y2+3y-13=-333+32+9-13+1-3=-32+32+9-13-1+3=11-13=323 sq. units

Page No 21.61:

Question 2:

Find the area bounded by the parabola x = 8 + 2yy2; the y-axis and the lines y = −1 and y = 3.

Answer:



The parabola cuts y-axis at (0, 4) and (0, −2).
Also, the points of intersection of the parabola and the lines y = 3 and y = −1 are B(5, 3) and D(5, −1) respectively.
Therefore, the area of the required region ABCDE
A=-13x dy=-138+2y-y2dy=8y+y2-y33-13=83+32-333-8-1+-12--133=24+9-9--8+1+13=24--7+13=24+7-13=31-13=923 sq. units

Page No 21.61:

Question 3:

Find the area bounded by the parabola y2 = 4x and the line y = 2x − 4.
(i) By using horizontal strips
(ii) By using vertical strips.

Answer:



To find the points of intersection between the parabola and the line let us substitute y = 2x − 4 in y2 = 4x.
2x-42=4x4x2+16-16x=4x4x2-20x+16=0x2-5x+4=0x-1x-4=0x=1, 4
y=-2, 4
Therefore, the points of intersection are C(1, −2) and A(4, 4).

(i) Using Horizontal Strips:
The area of the required region ABCD
 A=-24x1-x2dy            where, x1=y+42 and x2=y24=-24y+42-y24 dy=y24+2y-y312-24=424+2×4-4312--224+2-2--2312=4+8-163-1-4+23=12-163+3-23=15-183=15-183=15-6=9 sq. units


(ii) Using Vertical Strips:
The area of the required region ABCD
A=04y2dx-14y1dx          Where, y2=2x and y1=2x-4
 =042xdx-142x-4dx=43x3204-x2-4x14=43432-43032-42-4×4-12-4×1=323-0-0-1-4=323-3=233 square units

Page No 21.61:

Question 4:

Find the area of the region bounded by the parabola y2 = 2x and the straight line xy = 4.                    [NCERT EXEMPLAR]

Answer:


The parabola y2 = 2x opens towards the positive x-axis and its focus is 12,0.

The straight line xy = 4 passes through (4, 0) and (0, −4).

Solving y2 = 2x and xy = 4, we get

y2=2y+4y2-2y-8=0y-4y+2=0y=4 or y=-2

So, the points of intersection of the given parabola and the line are A(8, 4) and B(2, −2).



∴ Required area = Area of the shaded region OABO

                          =-24xlinedy--24xparabolady=-24y+4dy--24y22dy=y+422-24-12×y33-24=1264-4-1664--8=30-12=18 square units



Page No 21.62:

Question 1:

If the area above the x-axis, bounded by the curves y = 2kx and x = 0, and x = 2 is 3loge 2, then the value of k is
(a) 1/2
(b) 1
(c) −1
(d) 2

Answer:

(b) 1
The area bounded by the curves y=2kxx=0, and x=2 is given by 022kxdx.
It is given that 022kxdx=3loge2

 1k2kxloge202=3loge21k2k2loge2-2k0loge2=3loge21k22kloge2-1loge2=3loge21k22k-1=322k-1=3k22k-3k-1=0k=1

Clearly, k = 1 satisfies the equation.Hence, k = 1

Page No 21.62:

Question 2:

The area included between the parabolas y2 = 4x and x2 = 4y is (in square units)
(a) 4/3
(b) 1/3
(c) 16/3
(d) 8/3

Answer:

(c) 16/3





We have,x=y24        .....1x2=4y        .....2
Points of intersection of two parabola is given by,
y242=4yy4-64y=0yy3-64=0y=0, 4x=0, 4

Therefore, the points of intersection are A(0, 0) and C(4, 4).

Therefore, the area of the required region ABCD,

 =04y1-y2dx          where, y1=2x and y2=x24=042x-x24 dx=2×2x323-x31204=2×24323-4312-2×20323-0312=323-163-0=163 square units

Page No 21.62:

Question 3:

The area bounded by the curve y = loge x and x-axis and the straight line x = e is

(a) e sq. units

(b) 1 sq. units

(c) 1−1e sq. units

(d) 1+1e sq. units

Answer:

(b) 1 sq. units



The point of intersection of the curve and the straight line is A(e, 1).
Therefore, the area of the required region ABC,
 A=01x1-x2 dy             where, x1=e and x2=ey=01e-eydy=ey-ey01=e1-e1-e0-e0=e-e+1=1   square unit

Page No 21.62:

Question 4:

The area bounded by y = 2 − x2 and x + y = 0 is
(a) 72sq. units

(b) 92sq. units

(c) 9 sq. units

(d) none of these

Answer:

b  92 sq. units



To find the points of intersection of x + y = 0 and y = 2 − x2.
We put x = − y in y = 2 − x2, we get

y=2-y2y2+y-2=0y-1y+2=0y=1, -2
x=-1, 2
Therefore, the points of intersection are A(−1, 1) and C(2, −2).

The area of the required region ABCD,

 A=-12y1-y2dx        Where, y1=2-x2 and y2=-x=-122-x2+x dx=2x-x33+x22-12=22-233+222-2-1--133+-122=4-83+2--2+13+12=6-83+2-13-12=8-93-12=5-12=92  square units.

Page No 21.62:

Question 5:

The area bounded by the parabola x = 4 − y2 and y-axis, in square units, is
(a) 332

(b) 323

(c) 332

(d) 163

Answer:

b  323



The points of intersection of the parabola and the y-axis are A(0, 2) and C(0, −2).
Therefore, the area of the required region ABCO,
 A=-22x dy=-224-y2 dy=4y-y33-22=42-233-4-2--233=8-83--8+83=8-83+8-83=16-163=323 square units

Page No 21.62:

Question 6:

If An be the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x = π/4, then for x > 2
(a) An + An −2 = 1n-1

(b) An + An − 2 < 1n-1

(c) AnAn − 2 = 1n-1

(d) none of these

Answer:

(a) An + An −2 = 1n-1
An=Area bounded by the curve y=tanxn=tannx and the lines x=0, y=0, and x=π4.

Therefore,
 
An=0π4tannxdxAn-2=0π4tann-2xdx
Consider, An=0π4tannxdx
An=0π4tann-2xtan2xdxAn=0π4tann-2xsec2x-1dxAn=0π4tann-2xsec2x-tann-2xdxAn=0π4tann-2xsec2xdx-0π4tann-2xdx
An+An-2=0π4tann-2xsec2xdx

Now, An+An-2=0π4tann-2xsec2xdx

Let u=tanxdu=sec2xdx
Also, when x=0, u=0 and when x=π4, u=1

Therefore,

An+An-2=0π4tann-2xsec2xdx
=01un-2du=un-1n-101=1n-1-0=1n-1
 

Page No 21.62:

Question 7:

The area of the region formed by x2 + y2 − 6x − 4y + 12 ≤ 0, yx and x ≤ 5/2 is
(a) π6-3+18

(b) π6+3+18

(c) π6-3-18

(d) none of these

Answer:

c  π6-3-18



We have,x2+y2-6x-4y+120yxx52Following are the corresponding equations of the given inequation.x2+y2-6x-4y+12=0        .....1y=x                                     .....2x=52                                  .....3
Here, ABC is our required region in which point A is intersection of (1) and (3), point B is intersection of (1) and (2) and point C is intersection of (2) and (3).
By solving (1), (2) and (3) we get the coordinates of B and C as
B2, 2C52, 52
Now, the equation of the circle is,
x2+y2-6x-4y+12=0x-32+y-22=1y-22=1-x-32y-2=±1-x-32y=±1-x-32+2y=1-x-32 +2  or  -1-x-32+ 2y=1-x-32 +2 is not possible,Therefore, y=-1-x-32+ 2
The area of the required region ABC,
 A=252y2-y1 dx     Where, y1=-1-x-32+2 and y2=x=252x--1-x-32+2 dx=252x+1-x-32-2 dx=x22+x-321-x-32+12sin-1x-3-2x252=5222+52-321-52-32+12sin-152-3-252-222+2-321-2-32+12sin-12-3-22=258-141-14+12sin-1-12-5-2-12×0+12sin-1-1-4=-158-38+12×-π6-+12×-π2-2=-158-38-π12+π4+2=π6-3-18

Page No 21.62:

Question 8:

The area enclosed between the curves y = loge (x + e), x = loge 1y and the x-axis is
(a) 2
(b) 1
(c) 4
(d) none of these

Answer:

(a) 2



The point of intersection of the curves y=logex+e and x=loge1y is (0, 1)
y=logex+ex+e=eyx=ey-eHere taking, x1=ey-eand x2=loge1y
Therefore, area of the required region,
A=01x2-x1dy               Where, x1=ey-e and x2=loge1yA=01loge1ydy-01ey-edyA=01loge1ydy-ey-ey01     .....1Let I=loge1ydyPutting 1y=t-1y2dy=dtdy=-y2 dtdy=-1t2 dtTherefore, integral becomesI=-1t2 loget dt=-loge t 1t2dt-1t×1tdt=1 tloge t+1t=y loge 1y+yNow, 1 becomesA=y loge 1y+y01-ey-ey01 =y loge1y+y-ey+ey01=loge1+1-e1+e1-0+0-e0+e0=2

Page No 21.62:

Question 9:

The area of the region bounded by the parabola (y − 2)2 = x − 1, the tangent to it at the point with the ordinate 3 and the x-axis is
(a) 3
(b) 6
(c) 7
(d) none of these

Answer:



The tangent passes through the point with ordinate 3, so substituting y = 3 in equation of parabola (y − 2)2 = x − 1, we get x = 2
Therefore, the line touches the parabola at (2, 3).
We have,
y-22=x-1y-2=x-1y=x-1+2
Slope of the tangent of parabola at x = 2
dydxx=2=12x-1x=2=12
Therefore, the equation of the tangent is given as:
y-y0=mx-x0y-3=12x-2y=12x+2
Therefore, area of the required region ABC,
  A=03x1-x2dy            Where, x1=y-22+1 and x2=2y-2=03x1-x2dy=03y-22+1-2y-2dy=03y-2-12dy=03y-32dy=y-33303=3-333-0-333=9

Therefore the answer is (d)

Page No 21.62:

Question 10:

The area bounded by the curves y = sin x between the ordinates x = 0, x = π and the x-axis is
(a) 2 sq. units
(b) 4 sq. units
(c) 3 sq. units
(d) 1 sq. units

Answer:

(a) 2 sq. units



The required area ABC,
A=0πy dx=0πsinxdx=-cosx0π=-cosπ+cos0=1+1=2 square units

Page No 21.62:

Question 11:

The area bounded by the parabola y2 = 4ax and x2 = 4ay is
(a) 8a33

(b) 16a23

(c) 32a23

(d) 64a23

Answer:

b  16a23


To find the point of intersection of the parabolas substitute y=x24a in y2=4ax we get
x416a2=4axx4-64a3x=0xx3-64a3=0x=0  or  x=4ay=0  or  y=4a
Therefore, the required area ABCD,
   A=04ay1-y2dx      Where, y1=2ax  and  y2=x24a=04a2ax-x24adx=4a3x32-x312a04a=4a34a32-4a312a-4a3032-0312a=4a38a32-64a312a-0=32a23-16a23=16a23 square units

Page No 21.62:

Question 12:

The area bounded by the curve y = x4 − 2x3 + x2 + 3 with x-axis and ordinates corresponding to the minima of y is
(a) 1

(b) 9130

(c) 309

(d) 4

Answer:

b  9130



Clearly, from the figure the minimum value of y is 3 when x = 0 or 1.
Therefore, the required area ABCD,
 A=01y dx         Where, y=x4-2x3+x2+3=01x4-2x3+x2+3dx=x55-2x44+x33+3x01=155-2144+133+31-055-2044+033+30=15-12+13+3-0=6-15+10+9030=9130 square units



Page No 21.63:

Question 13:

The area bounded by the parabola y2 = 4ax, latusrectum and x-axis is
(a) 0

(b) 43a2

(c) 23a2

(d) a23

Answer:

b  43a2


Clearly, the latusrectum passes x-axis through the point D(a, 0).
Therefore, the required area ABCD,
   A=0ay dx        Where, y=2ax=012ax dx=4a3x320a=4a3a32-4a3032=43a2  square units

Page No 21.63:

Question 14:

The area of the region x,y : x2+y21x+y is
(a) π5

(b) π4

(c) π2-12

(d) π22

Answer:

Non of the given option is correct.



To find the points of intersection of the line and the circle substitute y = 1 − x in x2 + y2 = 1,we get A(0, 1) and B(1, 0).

Therefore, the required area of the shaded region,

 A=01y1-y2 dx      Where, y1=1-x2  and  y2=1-x=011-x2-1-x dx==011-x2-1+x dx=x21-x2+12sin-1x-x+x2201=121-12+12sin-11-1+122-021-02+12sin-10-0+022=π4-12 square units

Page No 21.63:

Question 15:

The area common to the parabola y = 2x2 and y = x2 + 4 is
(a) 23sq. units

(b) 32sq. units

(c) 323sq. units

(d) 332sq. units

Answer:

c 323  sq. units



Common region of two given parabola y = 2x2 and y = x2 + 4 is infinite as we see in the figure here.
Therefore, area common to these two parabola is infinity.

DISCLAIMER:
In the question, instead of
"The area common to the parabola y = 2x2 and y = x2 + 4 is"
It should be
"The closed area made by the parabola y = 2x2 and y = x2 + 4 is"
Solution of this question is as follow.



To find the point of intersection of the parabolas equate the equations y = 2x2 and y = x2 + 4 we get 

2x2=x2+4x2=4x=±2y=8

Therefore, the points of intersection are A(−2, 8) and C(2, 8).

Therefore, the required area ABCD,
 A=-22y1-y2dx       Where, y1=x2+4 and y2=2x2=-22x2+4-2x2 dx=-224-x2 dx=4x-x33-22=42-233-4-2--233=8-83--8+83=8-83+8-83=16-163=323  square units

Page No 21.63:

Question 16:

The area of the region bounded by the parabola y = x2 + 1 and the straight line x + y = 3 is given by
(a) 457

(b) 254

(c) π18

(d) 92

Answer:

d  92



To find the point of intersection of the parabola y = x2 + 1 and the line x + y = 3 substitute y = 3 − x in y = x2 + 1
3-x=x2+1x2+x-2=0x-1x+2=0x=1 or x=-2y=2 or y=5
So, we get the points of intersection A(−2, 5) and C(1, 2).
Therefore, the required area ABC,
 A=-21y1-y2dx          Where, y1=3-x and y2=x2+1=-213-x-x2+1dx=-213-x-x2-1dx=-212-x-x2dx=2x-x22-x33-21=21-122-133-2-2--222--233=2-12-13--4-2+83=2-12-13+4+2-83=8-12-93=5-12=92  square units

Page No 21.63:

Question 17:

The ratio of the areas between the curves y = cos x and y = cos 2x and x-axis from x = 0 to x = π/3 is
(a) 1 : 2
(b) 2 : 1
(c) 3 : 1
(d) none of these

Answer:

(d) none of these


The line x=π3 meets the curve y=cos x at Bπ3, 12
Area between the curve y = cos x and x-axis from x =0 and x = π3 is,
A1=0π3y1dx        Where,  y1=cosx=0π3cosxdx=sinx0π3=sinπ3-sin0=32



The line x=π3 meets the curve y=cos 2x at B'π3, -12

Area between the curve y = cos 2x and x-axis from x =0 and x = π3 is,
A2=0π4y2 dx-π4π3y2 dx               Where, y2 = cos 2x=0π4cos 2x dx-π4π3cos 2x dx=12sin 2x0π4-12sin 2xπ4π3=12sin π2-sin 0-12sin 2π3-sin π2=12-1232-1=12-34+12=1-34=4-34

Therefore the ratios will be

A1:A2=A1A2=324-34=234-3

Page No 21.63:

Question 18:

The area between x-axis and curve y = cos x when 0 ≤ x ≤ 2 π is
(a) 0
(b) 2
(c) 3
(d) 4

Answer:

(d) 4




Required shaded area,A=0π2cos x dx+π23π2-cos x dx+3π22πcos x dx=0π2cos x dx-π23π2cos x dx+3π22πcos x dx=sinx 0π2-sinx π23π2+sinx 3π22π=sinx 0π2-sinx π23π2+sinx 3π22π=1-0--1-1+0--1=1+2+1=4 sq units

Page No 21.63:

Question 19:

Area bounded by parabola y2 = x and straight line 2y = x is
(a) 43
(b) 1
(c) 23
(d) 13

Answer:


Disclaimer : Non of the given option is correct



Point of intersection is obtained by solving the equation of parabola y2 = x and equation of line 2y = x, we have
y2=x  and 2y =xy2=2y y2-2y =0y =0 or y=2x=0 or x=4Thus O0, 0 and A4, 2 are the points of intersection of the curve and straight line. Area bound by them A=04y1-y2 dx             Where, y1=x and y2=x2=04x-x2 dx=x3232-12×x2204=23x32-x2404=23432-14×42 -0=23×23-164=163-4=16-123=43  sq units

Page No 21.63:

Question 20:

The area bounded by the curve y = 4xx2 and the x-axis is
(a) 307sq. units

(b) 317sq. units

(c) 323sq. units

(d) 343sq. units

Answer:

c  323 sq. units


Point of intersection of parabola y = 4x − x2 with x-axis is given by
y=4x-x2 and y=0 Equation of x axis4x-x2=0x=0 or x=4 y=0 , y=0 Thus  O0, 0 and B4, 0  are points of intersection of parabola and x-axis.Required shaded area =044x -x2 dx                                =2 x2-x3304                                =2×16-643-0                                =96-643                                =323  square units 

Page No 21.63:

Question 21:

Area enclosed between the curve y2 (2ax) = x3 and the line x = 2a above x-axis is
(a) πa2

(b) 32πa2

(c) 2πa2

(d) 3πa2

Answer:



y2 2ax=x3y=x32ax
Let x = 2a sin2θ
    dx = 4a sinθ cosθ dθ

Area=02ax32axdx=0π28a3sin6θ2acos2θ·4asinθcosθdθ=8a20π2sin6θsinθdθ=8a20π2sin4θdθ=8a20π2sin2θ1cos2θdθ=8a20π21cos2θ2 dθ140π2sin22θ dθ=8a2 12θ0π2sin2θ40π2140π21cos 4θ2dθ=8a2 π4014π40=8a2 π4π16=32πa2

Page No 21.63:

Question 22:

The area of the region (in square units) bounded by the curve x2 = 4y, line x = 2 and x-axis is
(a) 1
(b) 2/3
(c) 4/3
(d) 8/3

Answer:

(b) 2/3



Point of intersection of the parabola x2 = 4y and straight line x = 2 is given by
x2=4y and x=24=4yy=1A2, 1 is the point of intersection of the curve and straight lineArea of shaded region OAB=02y dx                                       =02x24 dx                                        =x31202                                       =2312 -0                                       =23  square units 

Page No 21.63:

Question 23:

The area bounded by the curve y = f (x), x-axis, and the ordinates x = 1 and x = b is
(b −1) sin (3b + 4). Then, f (x) is
(a) (x − 1) cos (3x + 4)
(b) sin (3x + 4)
(c) sin (3x + 4) + 3 (x − 1) cos (3x +4)
(d) none of these

Answer:

(c) sin (3x + 4) + 3 (x − 1) cos (3x +4)

y=fx If A is the the area bound by the curve , x-axis , x=1 and x=b 1bfx dx =A1b=b-1sin 3b+4               Givenfx=ddxx-1sin3x+4=sin3x+4ddxx-1 +x-1ddxsin3x+4=sin3x+4+3x-1cos3x+4

Page No 21.63:

Question 24:

The area bounded by the curve y2 = 8x and x2 = 8y is
(a) 163sq. units

(b) 316sq. units

(c) 143sq. units

(d) 314sq. units

Answer:

Non of the given option is correct.



Point of intersection of both the parabolas y2 = 8x and x2 = 8y is obtained by solving the two equations

y2=8x and x2=8y  y464-8y=0yy3-83-0y=0  or  y=8x=0 or x=8 O0, 0 and A8, 8  are the points of intersection. Area of the shaded region =08y2-y1 dx                                      =08y2-y1dx                                      =088x-x28dx                                      =832x32-18×x3308                                      =23×8×832-18×833-0                                     =23×8×88 -823                                      =23×82-823                                      =8232-1                                      =643  sq units

Page No 21.63:

Question 25:

The area bounded by the parabola y2 = 8x, the x-axis and the latusrectum is
(a) 163

(b) 233

(c) 323

(d) 1623

Answer:

a  163




y2 = 8x represents a parabola opening side ways , with vertex at O(0, 0) and Focus at B(2, 0)

Thus AA' represents the latus rectum of the parabola.

The points of intersection of the parabola and latus rectum are  A(2, 4)  and A'(2, −4)

Area bound by curve , x-axis and latus rectum  is the area OABO,

The approximating rectangle of width =dx and  length=y has area =y dx, and moves from x=0 to x=2area OABO =02y dx=02y dx                      y>0 ,y=y=028xdx=2202xdx=22x323202=22×23232-0=423×22=163 sq units 

Page No 21.63:

Question 26:

Area bounded by the curve y = x3, the x-axis and the ordinates x = −2 and x = 1 is
(a) −9

(b) -154

(c) 154

(d) 174

Answer:

d  174



x=-2 and x=1 intersect the curve y = x3 at A-2, -8  and B 1, 1  respectivelyIf Px, y1  lies on OA  &  Q x, y2 lies on curve OB Then, y1>0 y1 =y1& y2<0y2 = -y2Area of curve bound by the two lines=shaded area OADO + Shaded area OCBO=-20y2 dx +01y1dx=-20-y2dx +01y1dx=-20-x3 dx +01x3 dx=-x44-20+x4401=0--164 +14-0=4+14=174 sq units



Page No 21.64:

Question 27:

The area bounded by the curve y = x |x| and the ordinates x = −1 and x = 1 is given by
(a) 0

(b) 13

(c) 23

(d) 43

Answer:

c  23



The given equation of the curve is
y= x xy = x2       x0-x2     x<0Now, solving x=1 and y=xx we getx=1  y=1 A1, 1  is point of intersection of the cuve y=xx and x=1Also, solving x=-1 and y=xx we getx=-1 y=-1A'-1, -1  is point of intersection of the cuve  y=xx and x=-1If Px, y1 , x>0  is a point on  y= x x   then y1>0   y1 =y1 And Qx, y2 , x<0  is a point on  y= x x   then y2<0  y2 = -y2Required area  =-10y2  dx +01y1 dx=-10-y2 dx +01y1 dx=-10--x2 dx +01x2 dx=-10x2 dx +01x2 dx=x33-10+x3301=0--133 +133-0=13+13=23  sq units 

Page No 21.64:

Question 28:

The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ xπ2 is
(a) 2 2-1

(b) 2-1

(c) 2+1

(d) 2

Answer:

(b) 2-1



Points of intersection is obtained by solving y=sinx and y=cos x  sin x= cos xx=π4 Thus the two functions intesect at x= π4  y=sin π4 =12Hence Aπ4 , 12  is the point of intersection.Area bound by the curves  and the y-axis when 0xπ2,A =012  x1  dy +121x2 dy=012 x1 dy +121x2dy=0 12sin-1y  dy +121cos-1 y dy=y sin-1 y +1-y2012+y cos -1y-1-y2121=12sin-1 12+1-12  -  1+1×cos -11-0-12cos -112+1-12  =    12×π4   +12 -1      +0-12×π4 +12=12+12-1=22-1=2-1  sq units 

Page No 21.64:

Question 29:

The area of the circle x2 + y2 = 16 enterior to the parabola y2 = 6x is
(a) 434π-3

(b) 434π+3

(c) 438π-3

(d) 438π+3

Answer:

(c) 438π-3



Points of intersection of the parabola and the circle is obtained by solving the simultaneous equations

x2+y2=16 and y2=6xx2+6x=16 x2+6x-16 =0x+8x-2=0x=2  or  x=-8 x can not be -8 as in this case it will be the point outside circle.x=2 When x=2, y=±6×2 =±12=±23B2 , 23 and B'2 , -23  are points of intersection of the parabola and circle. Required area = AreaOB'C'A'CBO=  area of circle-areaOBAB'O    Area of circle with radius 4 =π×42=16π         Now, AreaOBAB'O =2areaOBAO                       =2areaOBDO+areaDBAD                       =2×026x dx +2416-x2 dx                       =2×6x323202+x216-x2 +12×16 sin-1x424                      =2× 6×23×232-0 +12416-42 +12×16 sin-144-2216-22 -12×16 sin-124                      =2 ×  6×23×22+0+8 sin-11-12 -8 sin-112                      =2×833+8×π2-23-8π6                      =2 83-633+8π2-π6                      =2233+82π6                      =433+16π3Shaded area =16π-433+16π3=48π-16π3- 433=32π3- 433=438π-3  sq units

Page No 21.64:

Question 30:

Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is
(a) 2 (π − 2)
(b) π − 2
(c) 2π − 1
(d) 2 (π + 2)

Answer:

(b) π − 2



We have, x2 + y2 = 4 represents a circle with centre at O(0,0) and radius 2
x + y = 2 represents a straight line  cutting the x-axis at A(2, 0) and y axis at B(0, 2)
 
Thus , A (2,0) and B(0,2) are also the points of intersection of the straight line and the circle

Smaller area enclosed by the curve and straight line is the shaded area

Shaded area ABCA=areaOBCA - areaOBAO =024-x2 dx -022-xdx                    x2+y2=4y=4-x2  and  x+y=2y=2-x=024-x2 +x-2dx=12x4-x2 +12×4×sin-1x2 +x22-2x02=12×24-22 +2×sin-122 +222-2×2-0 =0+2×π2+2-4                       =π-2 sq units 

Page No 21.64:

Question 31:

Area lying between the curves y2 = 4x and y = 2x is
(a) 23

(b) 13

(c) 14

(d) 34

Answer:

b  13



The points of intersection of the straight line and the parabola is obtained by solving the simultaneous equations
y2=4x and y=2x2x2=4x4x2=4xxx-1=0x=0 or  x=1y=0  or y=2Thus O0, 0 and A1, 2  are the points of intersection of the parabola and straight lineShaded area is the required area.Using the horizontal strip method ,Shaded area =02x2-x1dy                  =02y2-y24 dy                 =12y22-14y3302                 =1422 -11223-0                =1-812                =12-812                =13  sq. units

Page No 21.64:

Question 32:

Area lying in first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2, is
(a) π

(b) π2

(c) π3

(d) π4

Answer:

(a) π



x2 + y2 = 4 represents a circle with centre at origin O(0, 0) and radius 2 units, cutting the coordinate axis at  A, A', B and B'.
x = 2 represents a straight line parallel to the y-axis, intersecting the circle at A(2, 0)
x = 0 represents the y-axis
Area bounded by the circle and the two given lines in the first quadrant is the shaded area OBCAO
AreaOBCAO=02y dx                    =024-x2 dx                    =12x4-x2 +12×4 sin-1x202                    =12×24-22 +12×4 sin-122 -0                    =0+2sin-11                    =2×π2                    =π  sq units 

Page No 21.64:

Question 33:

Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3, is
(a) 2

(b) 94

(c) 93

(d) 92

Answer:

b  94



y2 = 4x represents a parabola  with vertex at origin O(0, 0) and symmetric about +ve x-axis

y = 3 is a straight line parallel to the x-axis

Point of intersection of the line and the parabola is given by
Substituting y = 3  in the equation of the parabola
y2=4x32=4xx=94Thus A94 , 3 is the point of intersection of the parabola and straight line.

Required area is the shaded area OABO

Using the horizontal strip method , AreaOABO=03x dy                  =03y24 dy                 =14y3303                 =3312                 =94  sq. units



View NCERT Solutions for all chapters of Class 12