Class 12 Application of Integrals Mock Test (CBSE)

Topics in Application of Integrals

• Area under simple curves
• Area bounded by a curve and a line
• Area of regions bounded by lines and parabolas, circles and ellipses

Application of Integrals practice questions with answers

1. Q1Multiple choice·1 mark
   The area of the region bounded by the curve $y = \cos x$, the $x$-axis, and the ordinates $x = 0$ and $x = \frac{\pi}{2}$ is:

   1. (a)$1$
   2. (b)$2$
   3. (c)$\frac{\pi}{2}$
   4. (d)$0$
   Show answer and step marks

   1. Correct option: (a) $1$. Area = $\int_{0}^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1$.+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   The area of the region bounded by the circle $x^2 + y^2 = 4$ and the line $x = \sqrt{3}$ in the first quadrant is:

   1. (a)$\frac{\pi}{2} - \frac{\sqrt{3}}{2}$
   2. (b)$\frac{\pi}{6}$
   3. (c)$\frac{\pi}{3} - \frac{\sqrt{3}}{2}$
   4. (d)$\frac{2\pi}{3} - \frac{\sqrt{3}}{2}$
   Show answer and step marks

   1. Correct option: (c) $\frac{\pi}{3} - \frac{\sqrt{3}}{2}$. The area of the region is given by $\int_{\sqrt{3}}^{2} \sqrt{4 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{4-x^2} + 2\arcsin\left(\frac{x}{2}\right) \right]_{\sqrt{3}}^{2} = \pi - \left(\frac{\sqrt{3}}{2} + \frac{2\pi}{3}\right) = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$.+1

   Answer: (c)

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