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Class 12 RD Sharma Solutions – Chapter 21 Areas of Bounded Regions – Exercise 21.1 | Set 3

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  • Last Updated : 20 May, 2021
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Question 21. Draw a rough sketch of the curve y=\frac{π}{2}+2sin^2x  and find the area between x-axis, the curve and the ordinates x = 0, x = π

Solution:

Here, we have to find the bounded by 

y=\frac{\pi}{2}+2sin^2x

x-axis, x = 0 and x = π

Here is the table for values of y=\frac{\pi}{2}+2sin^2x

x  0\frac{\pi}{6}\frac{\pi}{4}                      \frac{\pi}{3}\frac{\pi}{2}\frac{2\pi}{3}                 \frac{3\pi}{4}\frac{5\pi}{6}π
\frac{\pi}{2}+2\ sin^2x  1.572.072.57                  3.073.573.07            2.57  2.071.57

Here is the rough sketch,

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = π,

Thus,

Required area = Region ABCDO

\displaystyle =\int_0^π y\ dx\\ =\int_0^π\left(\frac{π}{2}+2sin^2x\right)\ dx\\ =\int_0^π\left(\frac{π}{2}+1-cos\ 2x\right)\ dx\\ =\left[\frac{π}{2}x+x-\frac{sin\ 2x}{2}\right]_0^π\\ =\left[\left(\frac{π^2}{2}+π-\frac{sin\ 2x}{2}\right)-(0)\right]\\ =\frac{π^2}{2}+π

Required area = \frac{π}{2}(π+2)  square units

Question 22. Draw a rough sketch of the curve y=\frac{x}{π}+2sin^2x  and find the area between the x-axis, the curve and the ordinates x = 0, x = π.

Solution:

Here, we have the area between y-axis,

x = 0,

x = π

and

y=\frac{x}{π}+2\ sin^2x\ \ \ \ \ ......(1)

Thus, the table for equation (1) is

x0      \frac{π}{6}\frac{π}{4}\frac{π}{3}\frac{π}{2}\ \ \ \ \frac{2π}{3}\ \ \ \ \ \frac{3π}{4}\ \ \ \ \ \frac{5π}{6}π
y0         0.661.251.882.5  1.88   1.25   0.660

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y

Area of rectangle = y△x

The approx rectangles slide from x = 0 to x = π,

Thus,

Required area = Region ABOA

\displaystyle =\int_0^π y\ dx\\ =\int_0^π\left(\frac{π}{2}+2sin^2x\right)\ dx\\ =\int_0^π\left(\frac{π}{2}+1-cos\ 2x\right)\ dx\\ =\left[\frac{π}{2x}x+x-\frac{sin\ 2x}{2}\right]_0^π\\ =\left[\left(\frac{π^2}{2x}+π-0\right)-(0)\right]\\

Required area = \frac{3\pi}{2} square units

Question 23. Find the area bounded by the curve y = cos x between x = 0 and x = 2π

Solution:

Here from the figure we can see that

The required area = area of the region OABO + area of the region BCDB + area of the region DEFD

Therefore,

The required area = \displaystyle \int_0^{\frac{\pi}{2}}cos\ x\ dx+\left|\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}cos\ x\ dx\right|+\int_{\frac{3\pi}{2}}^{2\pi}cos\ x\ dx\\ =[sin\ x]_0^{\frac{\pi}{2}}+\left|[sin\ x]^{\frac{3\pi}{2}}_{\frac{\pi}{2}}\right|+[sin\ x]^{2\pi}_{\frac{3\pi}{2}}\\ =\left[sin\frac{\pi}{2}-sin0\right]+\left|sin\frac{3\pi}{2}-sin\frac{\pi}{2}\right|+\left[sin\ 2x-sin\frac{3\pi}{2}\right]\\ =1+2+1\\ =4\ sq.\ units

Question 24. Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x = \frac{π}{3}  are the ratio 2:3.

Solution:

We have to find the area under the curve

y = sin x    ……..(1)

and

y = sin 2x     …………(2)

Between x = 0 and x = \frac{\pi}{3}

x

y = sin x

0\ \ \ \ \ \ \frac{\pi}{6}\ \ \ \ \ \ \frac{\pi}{4}\ \ \ \ \ \ \ \frac{\pi}{3}\\ 0\ \ \ \ \ 0.5\ \ \ \ 0.7\ \ \ \ 0.8\frac{\pi}{2}\\ 1
y = sin 2x0      0.8       1       0.80

Here is the rough sketch 

Area under curve y = sin 2x

Shaded region represents the required area.

We slice it into approximation rectangle of 

Width = △x

Length = y1

Area of rectangle = y1△x

The approx rectangles slide from x = 0 to x = \frac{\pi}{3} ,

Thus,

Required area = Region OPACO

\displaystyle A_1=\int_0^{\frac{\pi}{3}}y_1\ dx\\ =\int_0^{\frac{\pi}{3}}sin\ 2x\ dx\\ =\left[\frac{-cos\ 2x}{2}\right]_0^{\frac{\pi}{3}}\\ =-\left[-\frac{1}{4}-\frac{1}{2}\right]\\ A_1=\frac{3}{4}\ sq.\ units

We slice it into approximation rectangle of 

Width = △x

Length = y2

Area of rectangle = y2△x

The approx rectangles slide from x = 0 to x = \frac{\pi}{3},

Thus,

Required area = Region OQACO

\displaystyle =\int_0^{\frac{\pi}{3}}y_2\ dx\\ =\int_0^{\frac{\pi}{3}}sin\ x\ dx\\ =\left[{-cos\ x}\right]_0^{\frac{\pi}{3}}\\ =-\left[\frac{1}{2}-1\right]\\ A_2=\frac{1}{2}\ sq.\ units

Thus,

A_2:A_1=\frac{1}{2}:\frac{3}{4}\\ A_2:A_1=2:3

Question 25. Compare the area under the curves y = cos2x and y = sin2x between x = 0 and x = π

Solution:

Here to compare area under curves

y = cos2x

and

y = sin2x

Between x = 0 and x = π

This is the table for y = cos2x and y = sin2x

x

y = cos2x

0

1

\frac{\pi}{6}\ \ \ \ \ \ \ \ \ \frac{\pi}{4}\\ 0.75\ \ \ \ 0.5\frac{\pi}{3}\ \ \ \ \ \ \ \ \ \frac{\pi}{2}\\ 0.25\ \ \ \ \ 0\frac{2\pi}{3}\ \ \ \ \ \ \ \ \ \frac{3\pi}{4}\\ 0.25\ \ \ \ \ \ 0.5\frac{5\pi}{6}\ \ \ \ \ \ \ \ \ \pi\\ 0.75\ \ \ \ \ 1
y = sin2x00.25          0.50.75      10.75     0.50.25       0

Area of region enclosed by

y = cos2x and axis

A1 = Region OABO + Region BCDB

= 2(Region BCDB)

\displaystyle =2\int_{\frac{\pi}{2}}^\pi cos^2x\ dx\\ =2\int_{\frac{\pi}{2}}^\pi\left(\frac{1-cos\ 2x}{2}\right)\ dx\\ =\left[x-\frac{sin\ 2x}{2}\right]^\pi_{\frac{\pi}{2}}\\ =\left[(x-0)-\left(\frac{\pi}{2}-0\right)\right]\\ =\pi-\frac{\pi}{2}\\ A_1=\frac{\pi}{2}\ sq.\ units\ \ \ \ \ \ .....(1)  

Area of region enclosed by y = sin2x and axis

A2 = Region OEDO

\displaystyle =\int_{0}^\pi sin^2x\ dx\\ =\int_{0}^\pi\left(\frac{1-cos\ 2x}{2}\right)\ dx\\ =\frac{1}{2}\left[x-\frac{sin\ 2x}{2}\right]^\pi_{0}\\ =\frac{1}{2}[(x-0)-(0)]\\ A_2=\frac{\pi}{2}\ sq.\ units\ \ \ \ \ \ .....(2)

From equation (1) and (2),

A1 = A2

Thus,

Area enclosed by y = cos2x = Area enclosed by y = sin2x.

Question 26. Find the area bounded by the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1  and the ordinates x = 0 and x = ae, where, b2 = a2(1 – e 2) and e < 1.

Solution:

Thus, the required area in the figure below of the region BOB’RFSB is enclosed by the ellipse and the lines x = 0 and x = ae

Here is the area of the region BOB’RFSB

\displaystyle =2\int_0^{ae}y\ dx\\ =2\frac{b}{a}\int_0^{ae}\sqrt{a^2-x^2}\ dx\\ =\frac{2b}{a}\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}\right]_0^{ae}\\ =\frac{2b}{2a}\left[ae\sqrt{a^2-a^2e^2}+a^2sin^{-1}e\right]\\ =ab\left[e\sqrt{1-e^2}+sin^{-1}e\right]

Question 27. Find the area of the minor segment of the circle x2 + y2 = a2 cut-off by the line x = \frac{a}{2}  .

Solution:

Area of the mirror segment of the circle

\displaystyle =2\int_{\frac{a}{2}}^a\sqrt{a^2-x^2}\ dx\\ =2\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{2}\right]_{\frac{a}{2}^a}\\ =2\left[\frac{a}{2}(0)+\frac{a^2}{2}sin^{-1}\left(\frac{a}{2}\right)-\frac{a}{4}\sqrt{a^2-\frac{a^2}{4}}-\frac{a^2}{2}sin^{-1}\frac{a}{4}\right]\\ =2\left[\frac{a^2}{2}sin^{-1}\left(\frac{a}{2}\right)-\frac{a}{4}\sqrt{a^2-\frac{a^2}{4}}-\frac{a^2}{2}sin^{-1}\frac{a}{4}\right]\\ =\frac{a^2}{12}[4\pi-3\sqrt3]\ sq.\ units

Question 28. Find the area of the region bounded by the curve x = at, y = 2at between the ordinates corresponding t = 1 and t = 2.

Solution:

Area of the bounded region

\displaystyle =2\int_1^2y\ \frac{dx}{dt}\ dt\\ =2\int_1^2(2at)(2at)\ dt\\ =8a^2\int_1^2t^2\ dt\\ =8a^2\left[\frac{t^3}{3}\right]_1^2\\ =8a^2\left[\frac{8}{3}-\frac{1}{3}\right]\\ =\frac{56a^2}{3}\ sq.\ units

Question 29. Find the area enclosed by the curve x = 3 cos t, y = 2 sin t.

Solution:

Area of the bounded region

\displaystyle =4\int_0^{\frac{\pi}{2}}2sin\ t\ dt\\ =-8[cos\ t]_0^{\frac{\pi}{2}}

= -8 [0 – 1]

= 8 square units 


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