₀² {(x+3)(2-x)} dx =

Correct Answer is : 258cos⁡−1(15)−6425/8 ^{-1}≤ft(1/5) - {{6}}{4}825cos−1(51)−46

25/8 ^{-1}≤ft(1/5) - {{6}}{4}

Correct Answer is : 258cos⁡−1(15)−6425/8 ^{-1}≤ft(1/5) - {{6}}{4}825cos−1(51)−46

Test Index

TG EAMCET 2 May 2025 Shift 1 Paper

Question : 76 of 160

Marks: +1, -0

\int\limits_{0}^{2} \sqrt{(x+3)(2-x)} dx =

Solution:

I\;=\int\limits_{0}^{2} \sqrt{(x+3)(2-x)} dx

\; \Rightarrow \int\limits_{0}^{2} \sqrt{-x^{2}-x+6} dx

\; \Rightarrow \int\limits_{0}^{2} \sqrt{-(x^{2}+x-6)} dx

\; \Rightarrow \int\limits_{0}^{2} \sqrt{\;\frac{25}{4} - \left(x+\frac{1}{2}\right)^{2}} dx

Put

u=x+\frac{1}{2} \Rightarrow du=dx

When

x=0, u=\frac{1}{2}

and

x=2

u\;=2+\frac{1}{2}=\frac{5}{2}

I\;=\int_{\frac{1}{2}}^{\frac{5}{2}} \sqrt{ \left(\frac{5}{2}\right)^{2} - u^{2} } du

\; \Rightarrow \left[ \frac{u}{2} \sqrt{\frac{25}{4} - u^{2}} + \frac{\frac{25}{4}}{2} \sin^{-1}\left( \frac{u}{\frac{5}{2}} \right) \right]_{\frac{1}{2}}^{\frac{5}{2}}

\; \Rightarrow \left[ \frac{u}{2} \sqrt{\frac{25}{4} - u^{2}} + \frac{25}{8} \sin^{-1}\left( \frac{2u}{5} \right) \right]_{\frac{1}{2}}^{\frac{5}{2}}

\; \Rightarrow \left\{ \frac{\frac{5}{2}}{2} \sqrt{\frac{25}{4} - \frac{25}{4}} + \frac{25}{8} \sin^{-1}(1) \right\} - \left\{ \frac{1}{4} \sqrt{\frac{25}{4} - \frac{1}{4}} + \frac{25}{8} \sin^{-1}\left(\frac{1}{5}\right) \right\}

\; \Rightarrow 0 + \frac{25}{8} \cdot \frac{\pi}{2} - \frac{1}{4} \sqrt{6} - \frac{25}{8} \sin^{-1}\left(\frac{1}{5}\right)

\; \Rightarrow \frac{25\pi}{16} - \frac{\sqrt{6}}{4} - \frac{25}{8} \sin^{-1}\left(\frac{1}{5}\right)

\; \Rightarrow \frac{25}{8} \left( \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{5}\right) \right) - \frac{\sqrt{6}}{4}

\; \Rightarrow \frac{25}{8} \cos^{-1}\left(\frac{1}{5}\right) - \frac{\sqrt{6}}{4}

Go to Question:

Prev Question

Next Question