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Mathematics · Definite Integration

JEE Advanced 2021 — Paper 2 — Question 47

Let gi:[π8,3π8]→R,i=1,2g_{i}:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow R, i=1,2 and f:[π8,3π8]→Rf:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow R be functions such that g1(x)=1,g2(x)=∣4x−π∣g_{1}(x)=1, g_{2}(x)=|4 x-\pi| and f(x)=sin⁡2xf(x)=\sin ^{2} x, for all x∈[π8,3π8]x \in\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right]. Define Si=∫π83π8f(x)⋅gi(x)dx,i=1,2S_{i}=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} f(x) \cdot g_{i}(x) d x, i=1,2

The value of 16S1π\frac{16 S_{1}}{\pi} is \qquad

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

& \mathrm{g}_{1}:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow \mathrm{R}, \mathrm{i}=1,2, \mathrm{f}:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow \mathrm{R} \\& \mathrm{~g}_{1}=1, \mathrm{~g}_{2}=|4 \mathrm{x}-\pi|, \mathrm{f}(\mathrm{x})=\sin ^{2} \mathrm{x} \\& \mathrm{~S}_{\mathrm{i}}=\int_{\pi / 8}^{3 \pi / 8} \mathrm{f}(\mathrm{x}) \cdot \mathrm{g}_{\mathrm{i}}(\mathrm{x}) \mathrm{dx} \\& \mathrm{~S}_{1}=\int_{\pi / 8}^{3 \pi / 8} \sin ^{2} \mathrm{xdx}=\int_{\pi / 8}^{3 \pi / 8} \sin ^{2}\left(\frac{\pi}{2}-\mathrm{x}\right) \mathrm{dx} \Rightarrow 2 \mathrm{~S}_{1}=\int_{\pi / 8}^{3 \pi / 8} 1 \mathrm{dx} \\& \Rightarrow \mathrm{~S}_{1}=\frac{1}{2}\left(\frac{3 \pi}{8}-\frac{\pi}{8}\right)=\frac{\pi}{8} \Rightarrow \frac{16 \mathrm{~S}_{1}}{\pi}=2 \end{aligned}$$

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Exam
JEE Advanced 2021
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration
Topic
Reduction Formulae in Definite Integrals

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