Definite Integration Question 3
Question 3 - 24 January - Shift 2
$\int _{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^{2}}} d x$ is equal to
(1) $\frac{\pi}{3}$
(2) $\frac{\pi}{2}$
(3) $\frac{\pi}{6}$
(4) $2 \pi$
Show Answer
Answer: (4)
Solution:
Formula: Standard formulas for Indefinite Integration
$\begin{aligned} & I=\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x \\ & \text { We have } \int \frac{d x}{\sqrt{a^2-x^2}}I=\sin ^{-1} \frac{x}{a}+C \\ & \text { Hence } \int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 \mathrm{x}^2}} \mathrm{dx}I=\frac{48}{2} \times\left[\sin ^{-1} \frac{2 \mathrm{x}}{3}\right]_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \\ & I=24 \times\left[\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{3}}{4}\right)-\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{2}}{4}\right)\right] \\ & I=24 \times\left[\sin ^{-1} \frac{\sqrt{3}}{2}-\sin ^{-1} \frac{1}{\sqrt{2}}\right] \\ & I=24 \times\left(\frac{\pi}{3}-\frac{\pi}{4}\right) \\ & I=24 \times \frac{\pi}{12}I=2 \pi \\ & \end{aligned}$
