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JEE Advance - Mathematics (2024 - Paper 2 Online - No. 17)

Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$.
Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$.
Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$.
The value of $\frac{16}{\pi^3} \int\limits_0^{\frac{\pi}{2}} f(x) g(x) d x$ is ______.
Answer
0.25

Explanation

Now $$I_1=\int_\limits0^{\frac{\pi}{2}} f(x) \cdot g(x) d x=\frac{1}{2} \int_\limits0^{\frac{\pi}{2}} g(x) d x$$

i.e. $$\frac{1}{2} \int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} d x$$

Using $$\int \sqrt{a^2-x^2}=\frac{1}{2}\left(x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)\right)+C$$

$$\begin{aligned} & \Rightarrow \frac{1}{2}\left[\frac{\left(x-\frac{\pi}{4}\right)}{2} \sqrt{\frac{\pi x}{2}-x^2}+\frac{\frac{\pi^2}{16}}{2} \sin ^{-1}\left(\frac{x-\frac{\pi}{4}}{\frac{\pi}{4}}\right)\right]_0^{\pi / 2} \\ & \Rightarrow \frac{1}{2}\left[\left(0+\frac{\pi^3}{64}\right)-\left(0+\left(\frac{-\pi^3}{64}\right)\right)\right] \\ & \Rightarrow \frac{1}{2} \times \frac{\pi^3}{32} \end{aligned}$$

Now $$\frac{16}{\pi^3} \times \frac{\pi^3}{64}=\frac{1}{4}=0.25$$

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