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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 30:

If the area between the curves  is divided into two equal parts by the line $x=a$, then find the value of a by using integration.

Curves  is divided into two equal parts by the line $x=a$

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 .

Ans

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:

Using integration, find the area of the region enclosed by the parabola $y=3{x}^{2}$ and the line $3x-y+6=0$.

Curves $y=3{x}^{2}$ and $y=3{x}^{2}+6$.