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Mathematics · Definite Integration

JEE Advanced 2021 — Paper 2 — Question 48

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 48 S2π2\frac{48 \mathrm{~S}_{2}}{\pi^{2}} is \qquad

Answer: 1.5

Numerical answer — enter this value.

Step-by-step solution

S2=∫π/83π/8sin⁡2x∣4x−π∣dx=∫π/83π/8sin⁡2(π2−x)∣ 4(π2−x)−π∣dx=∫π/83π/8cos⁡2x∣4x−π∣dx⇒2 S2=∫π/83π/8∣4x−π∣dx=2∫π/8π/4(π−4x)dx S2=(πx−2x2)∣π/8π/4=π(π4−π8)−2(π216−π264)S2=π28−3π232=π232⇒48 S2π2=48π2×π232=3248 S2π2=1.5\begin{aligned} & \begin{aligned} \mathrm{S}_{2}= & \left.\int_{\pi / 8}^{3 \pi / 8} \sin ^{2} x|4 \mathrm{x}-\pi| \mathrm{dx}=\int_{\pi / 8}^{3 \pi / 8} \sin ^{2}\left(\frac{\pi}{2}-\mathrm{x}\right) \right\rvert\, 4\left(\frac{\pi}{2}-\mathrm{x}\right)-\pi\rvert \mathrm{dx} \\& =\int_{\pi / 8}^{3 \pi / 8} \cos ^{2} x|4 \mathrm{x}-\pi| \mathrm{dx} \end{aligned} \\& \Rightarrow 2 \mathrm{~S}_{2}=\int_{\pi / 8}^{3 \pi / 8}|4 \mathrm{x}-\pi| \mathrm{dx}=2 \int_{\pi / 8}^{\pi / 4}(\pi-4 \mathrm{x}) \mathrm{dx} \\& \mathrm{~S}_{2}=\left.\left(\pi \mathrm{x}-2 \mathrm{x}^{2}\right)\right|_{\pi / 8} ^{\pi / 4}=\pi\left(\frac{\pi}{4}-\frac{\pi}{8}\right)-2\left(\frac{\pi^{2}}{16}-\frac{\pi^{2}}{64}\right) \\& \mathrm{S}_{2}=\frac{ \pi^{2}}{8}-\frac{3 \pi^{2}}{32}=\frac{\pi^{2}}{32} \Rightarrow\frac{48 \mathrm{~S}_{2}}{\pi^{2}}=\frac{48}{\pi^{2}} \times \frac{\pi^{2}}{32}=\frac{3}{2} \\& \frac{48 \mathrm{~S}_{2}}{\pi^{2}}=1.5 \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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