Let $\mathrm{g}_{i}:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow \mathbb{R}, \mathrm{i}=1,2$, and $f:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow \mathbb{R}$ be functions such that $\mathrm{g}_{1}(\mathrm{x})=1, \mathrm{~g}_{2}(\mathrm{x})=|4 \mathrm{x}-\pi|$ and $f(\mathrm{x})=\sin ^{2} \mathrm{x}$, for all $\mathrm{x} \in\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right]$. Define $\mathrm{S}_{i}=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} f(\mathrm{x}) \cdot \mathrm{g}_{i}(\mathrm{x}) \mathrm{dx}, i=1,2$. The value of $\frac{16 \mathrm{~S}_{1}}{\pi}$ is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Definite Integrals

Let

g_{i} : [\frac{π}{8}, \frac{3 π}{8}] \to R, i = 1, 2

, and

f : [\frac{π}{8}, \frac{3 π}{8}] \to R

be functions such that

g_{1} (x) = 1, g_{2} (x) = ∣4 x - π ∣

and

f (x) = sin^{2} x

, for all

x \in [\frac{π}{8}, \frac{3 π}{8}]

. Define

S_{i} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} f (x) \cdot g_{i} (x) dx, i = 1, 2

.

The value of

\frac{16 S _{1}}{π}

is ____.

Answer: 2

S_{1} = \int_{π /8}^{3 π /8} sin^{2} x d x

Using the property

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

S_{1} = \int_{π /8}^{3 π /8} sin^{2} (\frac{π}{2} - x) d x = \int_{π /8}^{3 π /8} cos^{2} x d x

Adding:

2 S_{1} = \int_{π /8}^{3 π /8} (sin^{2} x + cos^{2} x) d x = \int_{π /8}^{3 π /8} 1 d x = \frac{3 π}{8} - \frac{π}{8} = \frac{π}{4}

Therefore,

S_{1} = \frac{π}{8}

\frac{16 S _{1}}{π} = \frac{16 \times π /8}{π} = 2

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Numerical answer