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 $2 \int_{0}^{\frac{\pi}{2}} f(x) g(x) d x-\int_{0}^{\frac{\pi}{2}} g(x) d x$ is ____.

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Definite Integrals

Let

f : [0, \frac{π}{2}] \to [0, 1]

be the function defined by

f (x) = sin^{2} x

and let

g : [0, \frac{π}{2}] \to [0, \infty)

be the function defined by

g (x) = \frac{π x}{2} - x^{2}

.

The value of

2 \int_{0}^{\frac{π}{2}} f (x) g (x) d x - \int_{0}^{\frac{π}{2}} g (x) d x

is ____.

Answer: 0

Let

I_{1} = \int_{0}^{π /2} sin^{2} x \frac{π x}{2} - x^{2} d x

.

Rewrite

g (x) = (\frac{π}{4})^{2} - (x - \frac{π}{4})^{2}

.

Using King's property (substituting

x \to \frac{π}{2} - x

I_{1} = \int_{0}^{π /2} cos^{2} x (\frac{π}{4})^{2} - (\frac{π}{2} - x - \frac{π}{4})^{2} d x = \int_{0}^{π /2} cos^{2} x (\frac{π}{4})^{2} - (x - \frac{π}{4})^{2} d x

Adding both expressions:

2 I_{1} = \int_{0}^{π /2} (sin^{2} x + cos^{2} x) g (x) d x = \int_{0}^{π /2} g (x) d x

Therefore:

2 \int_{0}^{π /2} f (x) g (x) d x - \int_{0}^{π /2} g (x) d x = 2 I_{1} - \int_{0}^{π /2} g (x) d x = 0

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Numerical answer