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#### Question 1:

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

#### 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.

#### Question 3:

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

#### Question 4:

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

#### 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.

#### 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.

#### Question 7:

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

#### Question 8:

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

#### Question 9:

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

#### Question 10:

Draw a rough sketch of the graph of the curve $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$ and evaluate the area of the region under the curve and above the x-axis.

#### Question 11:

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

#### Question 12:

Draw a rough sketch of the graph of the function y = 2$\sqrt{1-{x}^{2}}$, x ∈ [0, 1] and evaluate the area enclosed between the curve and the x-axis.

#### Question 13:

Determine the area under the curve y = $\sqrt{{a}^{2}-{x}^{2}}$ included between the lines x = 0 and x = a.

#### 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.

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,

#### Question 15:

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

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.

#### 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.

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

Let the required area be A. Since limits on x are given, we use horizontal strips to find the area:

#### Question 17:

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

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

Integration represents the area enclosed by the graph from x = 0 to x = 1

#### Question 18:

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

We have,
y = | x + 3 | intersect x = 0 and x = −6 at (0, 3) and (−6, 3)
Now,

Integral represents the area enclosed between x = 6 and x = 0

#### Question 19:

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

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

Integral represents the area enclosed between x = −4 and x = 2

#### 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.

We have,
$xy-3x-2y-10=0\phantom{\rule{0ex}{0ex}}⇒xy-2y=3x+10\phantom{\rule{0ex}{0ex}}⇒y\left(x-2\right)=3x+10\phantom{\rule{0ex}{0ex}}⇒y=\frac{3x+10}{x-2}$
Let A represent the required area:

#### Question 21:

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

 x sin x $y=\frac{\mathrm{\pi }}{2}+2{\mathrm{sin}}^{2}x$ 0 $\frac{\mathrm{\pi }}{6}$ $\frac{\mathrm{\pi }}{2}$ $\frac{5\mathrm{\pi }}{6}$ $\mathrm{\pi }$ 0 $\frac{1}{2}$ 1 $\frac{1}{2}$ 0 1.57 2.07 3.57 2.07 1.57

#### Question 22:

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

The table for  different values of x and y is

 x sinx 0 $\frac{\mathrm{\pi }}{6}$ $\frac{\mathrm{\pi }}{2}$ $\frac{5\mathrm{\pi }}{6}$ $\mathrm{\pi }$ 0 $\frac{1}{2}$ 1 $\frac{1}{2}$ 0 0 $\frac{2}{3}$ $\frac{5}{2}$ $\frac{4}{3}$ 1

#### Question 23:

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

#### Question 24:

Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x = $\frac{\mathrm{\pi }}{3}$ are in the ratio 2 : 3.

#### Question 25:

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

Consider the value of  y  for different values of   x

 x 0 $\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}$ $\frac{\mathrm{\pi }}{3}$ $\frac{\mathrm{\pi }}{2}$ $\frac{2\mathrm{\pi }}{3}$ $\frac{5\mathrm{\pi }}{6}$ $\pi$ 1 0.5 0.25 0 0.25 0.75 1 0 0.5 0.75 1 0.75 0.25 0

Let A1 be the area of curve
Let A2 be the area of curve

Consider, a vertical strip of   length  $=\left|y\right|$   and width $=dx$ in the shaded region of both the curves

The area of approximating rectangle

#### Question 26:

Find the area bounded by the ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$ and the ordinates x = ae and x = 0, where b2 = a2 (1 − e2) and e < 1.

#### Question 27:

Find the area of the minor segment of the circle ${x}^{2}+{y}^{2}={a}^{2}$ cut off by the line $x=\frac{a}{2}$.

The equation of the circle is ${x}^{2}+{y}^{2}={a}^{2}$.

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

The line $x=\frac{a}{2}$ is parallel to y-axis and intersects the x-axis at $\left(\frac{a}{2},0\right)$.

Required area = Area of the shaded region

= 2 × Area of the region ABDA

$=2×{\int }_{\frac{a}{2}}^{a}{y}_{\mathrm{circle}}dx\phantom{\rule{0ex}{0ex}}=2{\int }_{\frac{a}{2}}^{a}\sqrt{{a}^{2}-{x}^{2}}dx\phantom{\rule{0ex}{0ex}}=2{\overline{)\left(\frac{x}{2}\sqrt{{a}^{2}-{x}^{2}}+\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}\frac{x}{a}\right)}}_{\frac{a}{2}}^{a}\phantom{\rule{0ex}{0ex}}=2\left[\left(0+\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}1\right)-\left(\frac{a}{4}×\sqrt{{a}^{2}-\frac{{a}^{2}}{4}}+\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}\frac{1}{2}\right)\right]$

#### Question 28:

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

The curve $x=a{t}^{2},y=2at$ represents the parametric equation of the parabola.

Eliminating the parameter t, we get

${y}^{2}=4ax$

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

#### Question 29:

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

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

Eliminating the parameter t, we get

$\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}={\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1$

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

∴ Required area = Area of the shaded region

= 4 × Area of the region OABO

#### 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.

#### Question 2:

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

#### Question 3:

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

#### Question 4:

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

#### Question 5:

Find the area of the region bounded by the curve $a{y}^{2}={x}^{3}$, the y-axis and the lines y = a and y = 2a.

The equation of the given curve is $a{y}^{2}={x}^{3}$.

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

#### Question 1:

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

#### Question 2:

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

#### Question 3:

Find the area of the region bounded by y = $\sqrt{x}$ and y = x.

#### Question 4:

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

#### Question 5:

Find the area of the region .

#### Question 6:

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

#### 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.

#### 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.

#### Question 9:

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

#### Question 10:

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

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

#### Question 11:

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

.......(1)       represents a circle with centre O(0,0) and radius  2
......(2)   represents a circle with centre A(2 ,0) and radius 2

Points of intersection  of two circles is given by solving the equations

#### Question 12:

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

We have, ${y}^{2}=x$ and $x+y=2$
To find the intersecting points of the curves ,we solve both the equations.

#### 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.

The  given region is the intersection of

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 $\sqrt{\frac{16}{3}}$.
Corresponding equations of the given inequations are

Substituting the value of y2 from (1) into (2)
$3{x}^{2}+9x=16$
$⇒3{x}^{2}+9x-16=0$
$⇒x=\frac{-9±\sqrt{81+192}}{6}$
$⇒x=\frac{-9±\sqrt{273}}{6}$
By figure we see that the value of x will be non-negative.
$\therefore x=\frac{-9+\sqrt{273}}{6}$
Now assume that x-coordinate of the intersecting point, $a=\frac{-9+\sqrt{273}}{6}$
The  Required area A  = 2(Area of OACO + Area of CABC)
Approximating the area of OACO  the length $=\left|{y}_{1}\right|$  width = dx
Area of   OACO

$=\frac{2\sqrt{3}{a}^{\frac{3}{2}}}{3}$
Therefore, Area of  OACO $=\frac{2\sqrt{3}{a}^{\frac{3}{2}}}{3}$
Similarly approximating the are of CABC the length $=\left|{y}_{2}\right|$   and the width = dx
Area of  CABC

$=\left[\frac{4}{2\sqrt{3}}\sqrt{\frac{16}{3}-\frac{16}{3}}+\frac{8}{3}{\mathrm{sin}}^{-1}\left(\frac{4\sqrt{3}}{4\sqrt{3}}\right)-\frac{a}{2}\sqrt{\frac{16}{3}-{a}^{2}}-\frac{8}{3}{\mathrm{sin}}^{-1}\left(\frac{a\sqrt{3}}{4}\right)\right]$

$=-\frac{a}{2}\sqrt{\frac{16}{3}-{a}^{2}}+\frac{8}{3}{\mathrm{sin}}^{-1}\left(1\right)-\frac{8}{3}{\mathrm{sin}}^{-1}\left(\frac{a\sqrt{3}}{4}\right)$
Area of  CABC $=-\frac{a}{2}\sqrt{\frac{16}{3}-{a}^{2}}+\frac{4\mathrm{\pi }}{3}-\frac{8}{3}{\mathrm{sin}}^{-1}\left(\frac{a\sqrt{3}}{4}\right)$
Thus the required area A = 2(Area of OACO + Area of CABC)
$=2\left[\frac{2\sqrt{3}{a}^{\frac{3}{2}}}{3}-\frac{a}{2}\sqrt{\frac{16}{3}-{a}^{2}}+\frac{4\mathrm{\pi }}{3}-\frac{8}{3}{\mathrm{sin}}^{-1}\left(\frac{a\sqrt{3}}{4}\right)\right]$
$=\frac{4{a}^{\frac{3}{2}}}{\sqrt{3}}-a\sqrt{\frac{16}{3}-{a}^{2}}+\frac{8\mathrm{\pi }}{3}-\frac{16}{3}{\mathrm{sin}}^{-1}\left(\frac{a\sqrt{3}}{4}\right)$

#### 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.

The given region is intersection of

Clearly, ${y}^{2}\le 5x$ is a parabola with vertex at origin and the axis is along the x-axis opening in the positive direction.Also $5{x}^{2}+5{y}^{2}\le 36$ is a circle with centre at the origin and has a radius .
Corresponding equations of given inequations are

Substituting the value of y2 from (1) into (2), we get

$5{x}^{2}+25x=36$
$⇒5{x}^{2}+25x-36=0$
$⇒x=\frac{-25±\sqrt{625+720}}{10}$

From the figure we see that x-coordinate of intersecting point can not be negative.
$\therefore x=\frac{-25+\sqrt{1345}}{10}$
Now assume that x-coordinate of intersecting point,  $a=\frac{-25+\sqrt{1345}}{10}$
The  Required area,
A  = 2(Area of OACO + Area of CABC)
Approximating the area of OACO  the length =$\left|{y}_{1}\right|$ and a width = dx=|y1=dx

= $\sqrt{5}{{\left[\frac{2{x}^{\frac{3}{2}}}{3}\right]}^{a}}_{0}$
Therefore, Area of OACO  $=\frac{2\sqrt{5}{a}^{\frac{3}{2}}}{3}$
Similarly approximating the area of CABC the length $=\left|{y}_{2}\right|$ and the width = dx

$={\int }_{a}^{\frac{6}{\sqrt{5}}}\sqrt{\frac{36}{5}-{x}^{2}}dx$

$=\frac{6}{2\sqrt{5}}\sqrt{\frac{36}{5}-\frac{36}{5}}+\frac{18}{5}{\mathrm{sin}}^{-1}\left(1\right)-\frac{a}{2}\sqrt{\frac{36}{5}-{a}^{2}}-\frac{18}{5}{\mathrm{sin}}^{-1}\left(\frac{a\sqrt{5}}{6}\right)$
$=0+\frac{18}{5}{\mathrm{sin}}^{-1}\left(1\right)-\frac{a}{2}\sqrt{\frac{36}{5}-{a}^{2}}-\frac{18}{5}{\mathrm{sin}}^{-1}\left(\frac{a\sqrt{5}}{6}\right)$
$=\frac{18}{5}×\frac{\mathrm{\pi }}{2}-\frac{a}{2}\sqrt{\frac{36}{5}-{a}^{2}}-\frac{18}{5}{\mathrm{sin}}^{-1}\left(\frac{a\sqrt{5}}{6}\right)$
Area of CABC  $=\frac{9\mathrm{\pi }}{5}-\frac{a}{2}\sqrt{\frac{36}{5}-{a}^{2}}-\frac{18}{5}{\mathrm{sin}}^{-1}\left(\frac{a\sqrt{5}}{6}\right)$
Thus the  Required area, A  = 2(Area of OACO + Area of CABC)

$=\frac{4\sqrt{5}{a}^{\frac{3}{2}}}{3}+\frac{18\mathrm{\pi }}{5}-a\sqrt{\frac{36}{5}-{a}^{2}}-\frac{36}{5}{\mathrm{sin}}^{-1}\left(\frac{a\sqrt{5}}{6}\right)$ ,         .

#### 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.

To find the points of intersection between two parabola let us substitute   $x=\frac{{y}^{2}}{4}$ in  ${x}^{2}=4y$.

Therefore, the points of intersection are  and .

Therefore, the area of the required region ABCD =${\int }_{0}^{4}{y}_{1}dx-{\int }_{0}^{4}{y}_{2}dx$ where ${y}_{1}=2\sqrt{x}$ and ${y}_{2}=\frac{{x}^{2}}{4}$

Required Area
$={\int }_{0}^{4}\left(2\sqrt{x}\right)dx-{\int }_{0}^{4}\left(\frac{{x}^{2}}{4}\right)dx\phantom{\rule{0ex}{0ex}}={\left[2×\frac{2{x}^{3}{2}}}{3}-\frac{{x}^{3}}{12}\right]}_{0}^{4}\phantom{\rule{0ex}{0ex}}=\left[\left(2×\frac{2{\left(4\right)}^{3}{2}}}{3}-\frac{{\left(4\right)}^{3}}{12}\right)-\left(2×\frac{2{\left(0\right)}^{3}{2}}}{3}-\frac{{\left(0\right)}^{3}}{12}\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

After simplifying we get,

$=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$ square units

#### Question 16:

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

To find the point of intersection of the parabolas substitute $y=\frac{{x}^{2}}{4b}$ in ${y}^{2}=4ax$ we get

Therefore, the required area ABCD${\int }_{0}^{4\sqrt[3]{a{b}^{2}}}\left({y}_{1}-{y}_{2}\right)\mathit{d}x$ where ${y}_{1}=2\sqrt{ax}$ and ${y}_{2}=\frac{{x}^{2}}{4b}$.

Required area =

#### Question 17:

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

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

represents a straight line passing through O(0,0)

Solving the two equations we get

#### Question 18:

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

The curve $y=\sqrt{x}$ or ${y}^{2}=x$ represents the parabola opening towards the positive x-axis.

The curve x = 2y + 3 represents a line passing through (3, 0) and $\left(0,-\frac{3}{2}\right)$.

Solving ${y}^{2}=x$ and x = 2y + 3, we get

∴ Required area = Area of the shaded region

#### 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}.

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

#### Question 20:

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

The given equations are and
Clearly the equation ${x}^{2}+{y}^{2}=8x$   is a circle with centre $\left(4,0\right)$ and has a radius 4.Also ${y}^{2}=4x$ 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
$\therefore$ ${x}^{2}+4x=8x$
$⇒$ ${x}^{2}-4x=0$
$⇒$$x\left(x-4\right)=0$
$⇒$
When
When

To approximate the area of the shaded region the length $=\left|{y}_{2}-{y}_{1}\right|$  and the width = dx
$A={\int }_{0}^{4}\left|{y}_{2}-{y}_{1}\right|dx$
$={\int }_{0}^{4}\left({y}_{2}-{y}_{1}\right)dx$

$=\left[0+0-0-8{\mathrm{sin}}^{-1}\left(\frac{-4}{4}\right)\right]-\frac{4}{3}×{4}^{\frac{3}{2}}$
$=\frac{8\mathrm{\pi }}{2}-\frac{32}{3}\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi }-\frac{32}{3}$
Hence the required area is   $4\mathrm{\pi }-\frac{32}{3}$  square units.

#### Question 21:

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

#### Question 22:

Prove that the area common to the two parabolas y = 2x2 and y = x2 + 4 is $\frac{32}{3}$ sq. units.

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

#### 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]

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

Equation of AB is

$y-5=\left(\frac{2-5}{-1-1}\right)\left(x-1\right)\phantom{\rule{0ex}{0ex}}⇒y-5=\frac{3}{2}\left(x-1\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{3}{2}x+5-\frac{3}{2}=\frac{3x+7}{2}$

Equation of BC is

$y-4=\left(\frac{5-4}{1-3}\right)\left(x-3\right)\phantom{\rule{0ex}{0ex}}⇒y-4=-\frac{1}{2}\left(x-3\right)\phantom{\rule{0ex}{0ex}}⇒y=-\frac{1}{2}x+4+\frac{3}{2}=\frac{-x+11}{2}$

Equation of CA is

$y-2=\left(\frac{4-2}{3+1}\right)\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒y-2=\frac{1}{2}\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{1}{2}x+2+\frac{1}{2}=\frac{x+5}{2}$

∴ Required area = Area of the shaded region

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

#### Question 24:

Find the area of the region bounded by $y=\sqrt{x}$ and y = x.                [NCERT EXEMPLAR]

The curve $y=\sqrt{x}$ or ${y}^{2}=x$ represents a parabola opening towards the positive x-axis.

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

Solving ${y}^{2}=x$ and y = x, we get

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

∴ Required area = Area of the shaded region OAO

#### Question 25:

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

represents a circle with centre O(0,0) and cutting the x axis at A(4,0)

represents straight passing through O(0,0)

Point of intersection is obtained by solving the two equations

#### Question 26:

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

We have,  and

To find the intersecting points of the curves ,we solve both the equations.

#### Question 27:

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

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

#### Question 28:

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

The curve $y=-{x}^{2}$ represents a parabola opening towards the negative y-axis.

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

Solving $y=-{x}^{2}$ and x + y + 2 = 0, we get

Thus, the parabola $y=-{x}^{2}$ 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

#### Question 29:

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

The graph of the parabola $y=2-{x}^{2}$ 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 $y=2-{x}^{2}$.

Therefore, the points of intersection are  and .

The area of the required region ABCD =${\int }_{-1}^{2}{y}_{1}\mathit{d}x-{\int }_{-1}^{2}{y}_{2}\mathit{d}x$ where ${y}_{1}=2-{x}^{2}$ and ${y}_{2}=-x$

Required Area
$={\int }_{-1}^{2}\left(2-{x}^{2}+x\right)\mathit{d}x\phantom{\rule{0ex}{0ex}}={\left[2x-\frac{{x}^{3}}{3}+\frac{{x}^{2}}{2}\right]}_{-1}^{2}\phantom{\rule{0ex}{0ex}}=\left[\left\{2\left(2\right)-\frac{{\left(2\right)}^{3}}{3}+\frac{{\left(2\right)}^{2}}{2}\right\}-\left\{2\left(-1\right)-\frac{{\left(-1\right)}^{3}}{3}+\frac{{\left(-1\right)}^{2}}{2}\right\}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

After simplifying we get,

$=\frac{9}{2}$ square units

#### 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.

We have,

#### 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.

We have, $y={x}^{2}+2$   and $y=x$
We see that parabola and the line $y=x$ do not intersect
is a line parallel to y axis

Point of intersection between parabola and  $x=1$ is

#### Question 32:

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

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

#### 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).

A(4, 1), B(6, 6) and C(8, 4) are three given points.

Required area of shaded region ABC

#### Question 34:

Using integration find the area of the region: .

$\left|x-1\right|\le y\le \sqrt{5-{x}^{2}}\phantom{\rule{0ex}{0ex}}\left|x-1\right|=\sqrt{5-{x}^{2}}\phantom{\rule{0ex}{0ex}}x=2,-1\phantom{\rule{0ex}{0ex}}\mathrm{A}={\int }_{-1}^{2}\left(\sqrt{5-{x}^{2}}-\left|x-1\right|\right)dx\phantom{\rule{0ex}{0ex}}={\int }_{-1}^{2}\sqrt{5-{x}^{2}}+{\int }_{-1}^{1}\left(x-1\right)dx+{\int }_{1}^{2}\left(1-x\right)dx\phantom{\rule{0ex}{0ex}}={\left[\frac{x}{2}\sqrt{5-{x}^{2}}+\frac{5}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{\sqrt{5}}\right)\right]}_{-1}^{2}+{\left[\frac{{x}^{2}}{2}-x\right]}_{-1}^{1}+{\left[x-\frac{{x}^{2}}{2}\right]}_{1}^{2}\phantom{\rule{0ex}{0ex}}=\frac{5}{2}\left({\mathrm{sin}}^{-1}\left(\frac{2}{\sqrt{5}}\right)+{\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)+\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 35:

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

#### 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.

We have, and $y=x$

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

#### Question 37:

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

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

#### Question 38:

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

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

#### 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.

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

#### Question 40:

Find the area of the region bounded by the curve y = $\sqrt{1-{x}^{2}}$, line y = x and the positive x-axis.

Hence, $y=\sqrt{1-{x}^{2}}$  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

#### Question 41:

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

All the three equations represent equations of straight lines
The points of intersection is obtained by solving  simultaneous equations

#### Question 42:

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

Let the two curves be named as y1 and y2 where

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

Therefore, intersection points are .
Now, the required area(OABO) =2[area(OACO) +area(CABC)]
Here,
$\mathrm{Area}\left(OACO\right)={\int }_{0}^{\frac{3}{2}}{Y}_{1}dx$
$={\int }_{0}^{\frac{3}{2}}\sqrt{9-{\left(x-3\right)}^{2}}dx$
And
$\mathrm{Area}\left(CABC\right)={\int }_{\frac{3}{2}}^{3}\left|{Y}_{2}\right|dx$
$={\int }_{\frac{3}{2}}^{3}\sqrt{9-{x}^{2}}dx$
Thus the required area is given by,
A = 2[area(OACO) +area(CABC)]

$=2{{\left[\frac{\left(x-3\right)}{2}\sqrt{9-{\left(x-3\right)}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\left(\frac{x-3}{3}\right)\right]}_{0}}^{\frac{3}{2}}+2{{\left[\frac{x}{2}\sqrt{9-{x}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{3}\right)\right]}^{3}}_{\frac{3}{2}}$

$=2\left[-\frac{9\sqrt{3}}{8}-\frac{9\mathrm{\pi }}{12}+\frac{9\mathrm{\pi }}{4}\right]+2\left[\frac{9\mathrm{\pi }}{4}-\frac{9\sqrt{3}}{8}-\frac{9\mathrm{\pi }}{12}\right]$

$=-\frac{18\sqrt{3}}{8}-\frac{18\mathrm{\pi }}{12}+\frac{18\mathrm{\pi }}{4}+\frac{18\mathrm{\pi }}{4}-\frac{18\sqrt{3}}{8}-\frac{18\mathrm{\pi }}{12}$

$=\frac{-36\sqrt{3}}{8}-\frac{36\mathrm{\pi }}{12}+\frac{36\mathrm{\pi }}{4}$

$=-\frac{9\sqrt{3}}{2}-3\mathrm{\pi }+9\mathrm{\pi }$
$=6\mathrm{\pi }-\frac{9\sqrt{3}}{2}$
Hence the required area is $6\mathrm{\pi }-\frac{9\sqrt{3}}{2}$  square units.

#### Question 43:

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

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 $\left|{y}_{2}-{y}_{1}\right|$ 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,
$A={\int }_{0}^{2}\left({y}_{2}-{y}_{1}\right)dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{2}\left[\sqrt{4-{x}^{2}}-\left(2-x\right)\right]dx\phantom{\rule{0ex}{0ex}}={\left[\frac{1}{2}x\sqrt{4-{x}^{2}}+\frac{4}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{2}\right)\right]}_{0}^{2}-{\left[2x-\frac{{x}^{2}}{2}\right]}_{0}^{2}\phantom{\rule{0ex}{0ex}}=\left[\frac{2}{2}\sqrt{4-{2}^{2}}+\frac{4}{2}{\mathrm{sin}}^{-1}\left(\frac{2}{2}\right)-\frac{0}{2}\sqrt{4-{0}^{2}}-\frac{4}{2}{\mathrm{sin}}^{-1}\left(\frac{0}{2}\right)\right]-\left[2\left(2\right)-\frac{{2}^{2}}{2}-2\left(0\right)+\frac{{0}^{2}}{2}\right]\phantom{\rule{0ex}{0ex}}=\left[0+2{\mathrm{sin}}^{-1}\left(1\right)-0-0\right]-\left[4-2-0+0\right]\phantom{\rule{0ex}{0ex}}=2{\mathrm{sin}}^{-1}\left(1\right)-2\phantom{\rule{0ex}{0ex}}=2×\frac{\mathrm{\pi }}{2}-2\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }-2$

#### Question 44:

Using integration, find the area of the following region: .

#### Question 45:

Using integration find the area of the region bounded by the curves and the x-axis.

The given curves are $y=\sqrt{4-{x}^{2}}$ and ${x}^{2}+{y}^{2}-4x=0$.
.....(1)
This represents a circle with centre O(0, 0) and radius = 2 units.
Also,
${x}^{2}+{y}^{2}-4x=0⇒{\left(x-2\right)}^{2}+{y}^{2}=4$         .....(2)
This represents a circle with centre B(2, 0) and radius = 2 units.
Solving (1) and (2), we get

Thus, the given circles intersect at and .

∴ Required area
= Area of the shaded region OABO

#### Question 46:

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

The given curves are

Clearly $y=\left|x-1\right|$ is cutting the x-axis  at (1, 0) and the y-axis at (0, 1) respectively.
Also $y=-\left|x-1\right|+1$ is cutting both the axes at (0, 0) and x-axis at (2, 0).
We have,

Thus  the intersecting points are  and .
The required area A = ( Area of  ABFA  + Area of  BCFB)
Now approximating the area of ABFA  the length = $\left|{y}_{1}\right|$   and width  = dx
Area of  ABFA
$={\int }_{1}{2}}^{1}\left[x-\left(1-x\right)\right]\mathit{d}x\phantom{\rule{0ex}{0ex}}={\int }_{1}{2}}^{1}\left(2x-1\right)\mathit{d}x\phantom{\rule{0ex}{0ex}}={\left[{x}^{2}-x\right]}_{1}{2}}^{1}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}$
Similarly approximating the area of  BCFB the length $=\left|{y}_{2}\right|$   and  width= dx
Area of  BCFB

Thus the required area A =( Area of ABFA  + Area of BCFB)
$=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
Hence the required area is $\frac{1}{2}$ square units.

#### Question 47:

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

represents a downward opening parabola, symmetrical about negative x-axis

#### Question 48:

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

We have, $y=4x-{x}^{2}$  and $y={x}^{2}-x$
The points of intersection of two curves is obtained by solving the simultaneous equations

#### Question 49:

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

We have, $y=4x-{x}^{2}$  and $y={x}^{2}-x$
The points of intersection of two curves is obtained by solving the simultaneous equations

#### Question 50:

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

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 − 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

#### Question 51:

If the area bounded by the parabola ${y}^{2}=4ax$ and the line y = mx is $\frac{{a}^{2}}{12}$ sq. units, then using integration, find the value of m.

The parabola ${y}^{2}=4ax$ opens towards the positive x-axis and its focus is (a, 0).

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

Solving ${y}^{2}=4ax$ and y = mx, we get

So, the points of intersection of the given parabola and line are O(0, 0) and $\mathrm{A}\left(\frac{4a}{{m}^{2}},\frac{4a}{m}\right)$.

∴ Area bounded by the given parabola and line

= Area of the shaded region

But,
Area bounded by the given parabola and line = $\frac{{a}^{2}}{12}$ sq. units          (Given)
$\therefore \frac{8{a}^{2}}{3{m}^{3}}=\frac{{a}^{2}}{12}\phantom{\rule{0ex}{0ex}}⇒{m}^{3}=32\phantom{\rule{0ex}{0ex}}⇒m=\sqrt[3]{32}$

Thus, the value of m is $\sqrt[3]{32}$.

#### Question 52:

If the area enclosed by the parabolas y2 = 16ax and x2 = 16ay, a > 0 is $\frac{1024}{3}$ square units, find the value of a.

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

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

But,

Area enclosed by the given parabolas = $\frac{1024}{3}$ square units                   (Given)

Thus, the value of a is 2.

#### Question 1:

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

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.

Therefore, the points of intersection are D(−1, −5) and A(3, 3).
The area of the required region ABCDOA,

#### Question 2:

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

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

#### 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.

To find the points of intersection between the parabola and the line let us substitute y = 2x − 4 in y2 = 4x.

Therefore, the points of intersection are C(1, −2) and A(4, 4).

(i) Using Horizontal Strips:
The area of the required region ABCD

(ii) Using Vertical Strips:
The area of the required region ABCD

#### Question 4:

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

The parabola y2 = 2x opens towards the positive x-axis and its focus is $\left(\frac{1}{2},0\right)$.

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

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

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

#### Question 1:

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

(b) 1
The area bounded by the curves $y={2}^{kx}$$x=0$, and $x=2$ is given by ${\int }_{0}^{2}{2}^{kx}dx$.
It is given that ${\int }_{0}^{2}{2}^{kx}dx=\frac{3}{{\mathrm{log}}_{e}\left(2\right)}$

$⇒\frac{1}{k}{\left[\frac{{2}^{kx}}{{\mathrm{log}}_{e}\left(2\right)}\right]}_{0}^{2}=\frac{3}{{\mathrm{log}}_{e}\left(2\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{k}\left[\frac{{2}^{k\left(2\right)}}{{\mathrm{log}}_{e}\left(2\right)}-\frac{{2}^{k\left(0\right)}}{{\mathrm{log}}_{e}\left(2\right)}\right]=\frac{3}{{\mathrm{log}}_{e}\left(2\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{k}\left(\frac{{2}^{2k}}{{\mathrm{log}}_{e}\left(2\right)}-\frac{1}{{\mathrm{log}}_{e}\left(2\right)}\right)=\frac{3}{{\mathrm{log}}_{e}\left(2\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{k}\left({2}^{2k}-1\right)=3\phantom{\rule{0ex}{0ex}}⇒{2}^{2k}-1=3k\phantom{\rule{0ex}{0ex}}⇒{2}^{2k}-3k-1=0\phantom{\rule{0ex}{0ex}}⇒k=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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

#### 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

(c) 16/3

Points of intersection of two parabola is given by,

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

Therefore, the area of the required region ABCD,

#### 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−$\frac{1}{e}$ sq. units

(d) 1+$\frac{1}{e}$ sq. units

(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,

#### Question 4:

The area bounded by y = 2 − x2 and x + y = 0 is
(a) $\frac{7}{2}$sq. units

(b) $\frac{9}{2}$sq. units

(c) 9 sq. units

(d) none of these

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

Therefore, the points of intersection are A(−1, 1) and C(2, −2).

The area of the required region ABCD,

#### Question 5:

The area bounded by the parabola x = 4 − y2 and y-axis, in square units, is
(a) $\frac{3}{32}$

(b) $\frac{32}{3}$

(c) $\frac{33}{2}$

(d) $\frac{16}{3}$

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,

#### 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 = $\frac{1}{n-1}$

(b) An + An − 2 < $\frac{1}{n-1}$

(c) AnAn − 2 = $\frac{1}{n-1}$

(d) none of these

(a) An + An −2 = $\frac{1}{n-1}$
${A}_{n}=$Area bounded by the curve $y={\left\{\mathrm{tan}\left(x\right)\right\}}^{n}={\mathrm{tan}}^{n}\left(x\right)$ and the lines and $x=\frac{\mathrm{\pi }}{4}$.

Therefore,

${A}_{n}={\int }_{0}^{\mathrm{\pi }}{4}}{\mathrm{tan}}^{n}\left(x\right)\mathit{d}x\phantom{\rule{0ex}{0ex}}⇒{A}_{n-2}={\int }_{0}^{\mathrm{\pi }}{4}}{\mathrm{tan}}^{n-2}\left(x\right)\mathit{d}x$
Consider, ${A}_{n}={\int }_{0}^{\mathrm{\pi }}{4}}{\mathrm{tan}}^{n}\left(x\right)\mathit{d}x$
$⇒{A}_{n}={\int }_{0}^{\mathrm{\pi }}{4}}\left\{{\mathrm{tan}}^{n-2}\left(x\right)\right\}\left\{{\mathrm{tan}}^{2}\left(x\right)\right\}\mathit{d}x\phantom{\rule{0ex}{0ex}}⇒{A}_{n}={\int }_{0}^{\mathrm{\pi }}{4}}\left\{{\mathrm{tan}}^{n-2}\left(x\right)\right\}\left\{{\mathrm{sec}}^{2}\left(x\right)-1\right\}\mathit{d}x\phantom{\rule{0ex}{0ex}}⇒{A}_{n}={\int }_{0}^{\mathrm{\pi }}{4}}\left\{{\mathrm{tan}}^{n-2}\left(x\right){\mathrm{sec}}^{2}\left(x\right)-{\mathrm{tan}}^{n-2}\left(x\right)\right\}\mathit{d}x\phantom{\rule{0ex}{0ex}}⇒{A}_{n}={\int }_{0}^{\mathrm{\pi }}{4}}\left\{{\mathrm{tan}}^{n-2}\left(x\right){\mathrm{sec}}^{2}\left(x\right)\right\}\mathit{d}x-{\int }_{0}^{\mathrm{\pi }}{4}}{\mathrm{tan}}^{n-2}\left(x\right)\mathit{d}x$
$⇒{A}_{n}+{A}_{n-2}={\int }_{0}^{\mathrm{\pi }}{4}}{\mathrm{tan}}^{n-2}\left(x\right){\mathrm{sec}}^{2}\left(x\right)\mathit{d}x$

Now, ${A}_{n}+{A}_{n-2}={\int }_{0}^{\mathrm{\pi }}{4}}{\mathrm{tan}}^{n-2}\left(x\right){\mathrm{sec}}^{2}\left(x\right)\mathit{d}x$

Also, when  and when

Therefore,

${A}_{n}+{A}_{n-2}={\int }_{0}^{\mathrm{\pi }}{4}}{\mathrm{tan}}^{n-2}\left(x\right){\mathrm{sec}}^{2}\left(x\right)\mathit{d}x$
$={\int }_{0}^{1}\left({u}^{n-2}\right)du\phantom{\rule{0ex}{0ex}}={\left[\frac{{u}^{n-1}}{n-1}\right]}_{0}^{1}\phantom{\rule{0ex}{0ex}}=\left[\frac{1}{n-1}-0\right]=\frac{1}{n-1}$

#### Question 7:

The area of the region formed by x2 + y2 − 6x − 4y + 12 ≤ 0, yx and x ≤ 5/2 is
(a) $\frac{\mathrm{\pi }}{6}-\frac{\sqrt{3}+1}{8}$

(b) $\frac{\mathrm{\pi }}{6}+\frac{\sqrt{3}+1}{8}$

(c) $\frac{\mathrm{\pi }}{6}-\frac{\sqrt{3}-1}{8}$

(d) none of these

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

Now, the equation of the circle is,

The area of the required region ABC,

#### Question 8:

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

(a) 2

The point of intersection of the curves  is (0, 1)

Therefore, area of the required region,

#### 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

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,
${\left(y-2\right)}^{2}=x-1\phantom{\rule{0ex}{0ex}}⇒y-2=\sqrt{x-1}\phantom{\rule{0ex}{0ex}}⇒y=\sqrt{x-1}+2$
Slope of the tangent of parabola at x = 2
${\left[\frac{\mathit{d}y}{\mathit{d}x}\right]}_{x=2}={\left[\frac{1}{2\sqrt{x-1}}\right]}_{x=2}=\frac{1}{2}$
Therefore, the equation of the tangent is given as:
$y-{y}_{0}=m\left(x-{x}_{0}\right)\phantom{\rule{0ex}{0ex}}⇒y-3=\frac{1}{2}\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{1}{2}x+2$
Therefore, area of the required region ABC,

#### 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

(a) 2 sq. units

The required area ABC,

#### Question 11:

The area bounded by the parabola y2 = 4ax and x2 = 4ay is
(a) $\frac{8{a}^{3}}{3}$

(b) $\frac{16{a}^{2}}{3}$

(c) $\frac{32{a}^{2}}{3}$

(d) $\frac{64{a}^{2}}{3}$

To find the point of intersection of the parabolas substitute $y=\frac{{x}^{2}}{4a}$ in ${y}^{2}=4ax$ we get

Therefore, the required area ABCD,

#### 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) $\frac{91}{30}$

(c) $\frac{30}{9}$

(d) 4

Clearly, from the figure the minimum value of y is 3 when x = 0 or 1.
Therefore, the required area ABCD,

#### Question 13:

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

(b) $\frac{4}{3}{a}^{2}$

(c) $\frac{2}{3}{a}^{2}$

(d) $\frac{{a}^{2}}{3}$

Clearly, the latusrectum passes x-axis through the point D(a, 0).
Therefore, the required area ABCD,

#### Question 14:

The area of the region is
(a) $\frac{\mathrm{\pi }}{5}$

(b) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{\mathrm{\pi }}{2}-\frac{1}{2}$

(d) $\frac{{\mathrm{\pi }}^{2}}{2}$

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,

#### Question 15:

The area common to the parabola y = 2x2 and y = x2 + 4 is
(a) $\frac{2}{3}$sq. units

(b) $\frac{3}{2}$sq. units

(c) $\frac{32}{3}$sq. units

(d) $\frac{3}{32}$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:
"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

$2{x}^{2}={x}^{2}+4\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=4\phantom{\rule{0ex}{0ex}}⇒x=±2\phantom{\rule{0ex}{0ex}}\therefore y=8$

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

Therefore, the required area ABCD,

#### 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) $\frac{45}{7}$

(b) $\frac{25}{4}$

(c) $\frac{\mathrm{\pi }}{18}$

(d) $\frac{9}{2}$

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

So, we get the points of intersection A(−2, 5) and C(1, 2).
Therefore, the required area ABC,

#### 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) $\sqrt{3}$ : 1
(d) none of these

(d) none of these

Area between the curve y = cos x and x-axis from x =0 and x = $\frac{\mathrm{\pi }}{3}$ is,

Area between the curve y = cos 2x and x-axis from x =0 and x = $\frac{\mathrm{\pi }}{3}$ is,

Therefore the ratios will be

${A}_{1}:{A}_{2}=\frac{{A}_{1}}{{A}_{2}}=\frac{\frac{\sqrt{3}}{2}}{\frac{4-\sqrt{3}}{4}}=\frac{2\sqrt{3}}{4-\sqrt{3}}$

#### 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

(d) 4

#### Question 19:

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

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

#### Question 20:

The area bounded by the curve y = 4xx2 and the x-axis is
(a) $\frac{30}{7}$sq. units

(b) $\frac{31}{7}$sq. units

(c) $\frac{32}{3}$sq. units

(d) $\frac{34}{3}$sq. units

Point of intersection of parabola y = 4x − x2 with x-axis is given by

#### Question 21:

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

(b) $\frac{3}{2}\mathrm{\pi }{a}^{2}$

(c) 2πa2

(d) 3πa2

Let x = 2a sin2θ
dx = 4a sinθ cosθ dθ

#### 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

(b) 2/3

Point of intersection of the parabola x2 = 4y and straight line x = 2 is given by

#### 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

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

#### Question 24:

The area bounded by the curve y2 = 8x and x2 = 8y is
(a) $\frac{16}{3}$sq. units

(b) $\frac{3}{16}$sq. units

(c) $\frac{14}{3}$sq. units

(d) $\frac{3}{14}$sq. units

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

#### Question 25:

The area bounded by the parabola y2 = 8x, the x-axis and the latusrectum is
(a) $\frac{16}{3}$

(b) $\frac{23}{3}$

(c) $\frac{32}{3}$

(d) $\frac{16\sqrt{2}}{3}$

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,

#### Question 26:

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

(b) $\frac{-15}{4}$

(c) $\frac{15}{4}$

(d) $\frac{17}{4}$

#### 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) $\frac{1}{3}$

(c) $\frac{2}{3}$

(d) $\frac{4}{3}$

The given equation of the curve is

#### Question 28:

The area bounded by the y-axis, y = cos x and y = sin x when 0 ≤ x$\frac{\mathrm{\pi }}{2}$ is
(a) 2 $\left(\sqrt{2}-1\right)$

(b) $\sqrt{2}-1$

(c) $\sqrt{2}+1$

(d) $\sqrt{2}$

(b) $\sqrt{2}-1$

#### Question 29:

The area of the circle x2 + y2 = 16 enterior to the parabola y2 = 6x is
(a) $\frac{4}{3}\left(4\mathrm{\pi }-\sqrt{3}\right)$

(b) $\frac{4}{3}\left(4\mathrm{\pi }+\sqrt{3}\right)$

(c) $\frac{4}{3}\left(8\mathrm{\pi }-\sqrt{3}\right)$

(d) $\frac{4}{3}\left(8\mathrm{\pi }+\sqrt{3}\right)$

(c) $\frac{4}{3}\left(8\mathrm{\pi }-\sqrt{3}\right)$

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

#### 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)

(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

#### Question 31:

Area lying between the curves y2 = 4x and y = 2x is
(a) $\frac{2}{3}$

(b) $\frac{1}{3}$

(c) $\frac{1}{4}$

(d) $\frac{3}{4}$

The points of intersection of the straight line and the parabola is obtained by solving the simultaneous equations

#### 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) $\frac{\mathrm{\pi }}{2}$

(c) $\frac{\mathrm{\pi }}{3}$

(d) $\frac{\mathrm{\pi }}{4}$

(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

#### Question 33:

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

(b) $\frac{9}{4}$

(c) $\frac{9}{3}$

(d) $\frac{9}{2}$

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

Required area is the shaded area OABO

View NCERT Solutions for all chapters of Class 15