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 $\frac{48 \mathrm{~S}_{2}}{\pi^{2}}$ is ____.

Definite Integrals — JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 Question with Solution

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{48 S _{2}}{π ^{2}}

is ____.

Answer: 1.5

S_{2} = \int_{π /8}^{3 π /8} sin^{2} x ∣4 x - π ∣ d x

Note that

4 x = π

when

x = π /4

, which is in the interval

[π /8, 3 π /8]

.

Using the substitution property

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

S_{2} = \int_{π /8}^{3 π /8} cos^{2} x ∣ π - 4 x ∣ d x = \int_{π /8}^{3 π /8} cos^{2} x ∣4 x - π ∣ d x

Adding:

2 S_{2} = \int_{π /8}^{3 π /8} ∣4 x - π ∣ d x

Since

∣4 x - π ∣ = π - 4 x

for

x \in [π /8, π /4]

and

∣4 x - π ∣ = 4 x - π

for

x \in [π /4, 3 π /8]

2 S_{2} = 2 \int_{π /8}^{π /4} (π - 4 x) d x = 2 \times \frac{1}{2} \times \frac{π}{8} \times \frac{π}{2} = \frac{π ^{2}}{16}

Therefore,

S_{2} = \frac{π ^{2}}{32}

\frac{48 S _{2}}{π ^{2}} = \frac{48 \times π ^{2} /32}{π ^{2}} = \frac{48}{32} = \frac{3}{2} = 1.5

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