CBSE Class 12 Mathematics Application of Integrals Class Test Set 01

Class Test Questions with Solutions

Question 1. Find the area of the region bounded by \( x^2 + y^2 = 8 \) in the first quadrant using integration. [1]
Answer:
Given circle is \( x^2 + y^2 = 8 \) ..... (1)

Required shaded area \( A = \int_{0}^{2\sqrt{2}} y\,dx \)
From equation (1): \( y = \sqrt{8 - x^2} \)
\( A = \int_{0}^{2\sqrt{2}} \sqrt{8 - x^2}\,dx \)
\( = \left[\frac{x}{2}\sqrt{8 - x^2} + \frac{8}{2}\sin^{-1}\frac{x}{2\sqrt{2}}\right]_{0}^{2\sqrt{2}} \)
\( = \left[0 + 4\sin^{-1}\frac{2\sqrt{2}}{2\sqrt{2}} - 0\right] \)
\( = 4 \times \frac{\pi}{2} = 2\pi \) sq. units

Question 2. Find the area of the region bounded by \( y = \sqrt{x} \), x = 2 and the x-axis. [1]
Answer:
Area bounded by \( y = \sqrt{x} \), x = 2 and by x-axis

Required shaded area \( A = \int_{0}^{2} y\,dx = \int_{0}^{2} \sqrt{x}\,dx \)
\( = \left[\frac{x^{3/2}}{3/2}\right]_{0}^{2} = \frac{2}{3}\left(x^{3/2}\right)_{0}^{2} \)
\( = \frac{2}{3}\left(2^{3/2} - 0\right) = \frac{2}{3} \cdot 2\sqrt{2} = \frac{4\sqrt{2}}{3} \) sq. units

Question 3. Area of the region enclosed by y = sin x, \( \frac{\pi}{2} \leq x \leq \frac{3\pi}{2} \) and x-axis is: [2]
Answer:
Area enclosed by y = sin x, \( \frac{\pi}{2} \leq x \leq \frac{3\pi}{2} \) and x-axis is

Required shaded area \( A = \int_{\pi/2}^{\pi} y\,dx + \left| \int_{\pi}^{3\pi/2} y\,dx \right| \)
\( = \int_{\pi/2}^{\pi} \sin x\,dx + \left| \int_{\pi}^{3\pi/2} \sin x\,dx \right| \)
\( = -\left[\cos x\right]_{\pi/2}^{\pi} + \left| \left[-\cos x\right]_{\pi}^{3\pi/2} \right| = -(-1 - 0) + \left| -(0 - (-1)) \right| \)
= 1 + 1 = 2 square units

Question 4. Find the area enclosed by \( x^2 = 2y \), y = 3 and the y-axis. [2]
Answer:
Area enclosed by \( x^2 = 2y \), y = 3 and the y-axis (the region lies on one side of the y-axis; the first-quadrant part is shown shaded)

Required shaded area \( A = \int_{0}^{3} x\,dy = \int_{0}^{3} \sqrt{2y}\,dy \)
\( = \sqrt{2}\left(\frac{y^{3/2}}{3/2}\right)_{0}^{3} = \frac{2\sqrt{2}}{3}\left(3^{3/2} - 0\right) \)
\( = \frac{2\sqrt{2}}{3}\left(3\sqrt{3}\right) = 2\sqrt{6} \) square units

Question 5. Find the area enclosed by ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \) and lying above the x-axis. [3]
Answer:
Area enclosed by ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \) and lying above the x-axis.
Put x = 0, y = ±3; put y = 0, x = ±2

Required shaded area \( A = 2\int_{0}^{2} y\,dx \) .....(1)
\( \frac{x^2}{4} + \frac{y^2}{9} = 1 \)
⇒ \( \frac{y^2}{9} = 1 - \frac{x^2}{4} \)
⇒ \( y = \frac{3}{2}\sqrt{2^2 - x^2} \), put in equation (1)
⇒ \( A = 2 \cdot \frac{3}{2}\int_{0}^{2} \sqrt{2^2 - x^2}\,dx \)
⇒ \( A = 3\left[\frac{x}{2}\sqrt{2^2 - x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2}\right]_{0}^{2} \)
⇒ \( A = 3\left[\frac{4}{2} \times \sin^{-1}\left(\frac{2}{2}\right) - 0\right] \)
\( = 3\left[2 \times \frac{\pi}{2} - 0\right] = 3\pi \) square units

Question 6. Find the area enclosed by lines x + 2y = 8, x = 2, x = 4 and x-axis. [3]
Answer:
Area enclosed by lines x + 2y = 8, x = 2, x = 4 and x-axis.
2y = −x + 8
\( y = \frac{-x}{2} + 4 \)
put x = 0, y = 4
put y = 0, x = 8

Required shaded area \( A = \int_{2}^{4} y\,dx = \int_{2}^{4} \left(-\frac{x}{2} + 4\right) dx \)
\( = \left(-\frac{x^2}{4} + 4x\right)_{2}^{4} \)
\( = \left[-\frac{16}{4} + 16 - \left(-\frac{4}{4} + 8\right)\right] \)
= [−4 + 16 − (−1 + 8)]
= [12 − 7] = 5 square units