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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2020 · Mathematics · Definite Integration

∫_ π / 3 ^ π / 2 x (π[x]-x) dx is equal to

∫π/3π/2xsin⁡(π[x]−x)dx\int_{\pi / 3}^{\pi / 2} x \sin (\pi[x]-x) \mathrm{dx} is equal to

Options

  1. A12+π6\frac{1}{2}+\frac{\pi}{6}
  2. B1−32+π61-\frac{\sqrt{3}}{2}+\frac{\pi}{6}
  3. C−12−π6-\frac{1}{2}-\frac{\pi}{6}
  4. D32−1−π6\frac{\sqrt{3}}{2}-1-\frac{\pi}{6}

Correct answer

B. 1−32+π61-\frac{\sqrt{3}}{2}+\frac{\pi}{6}

Step-by-step solution

In the interval π3\frac{\pi}{3} to π2,[x]=1\frac{\pi}{2},[\mathrm{x}]=1∴I=∫π/3π/2xsin⁡(π−x)dx=∫π/3π/2xsin⁡xdx=[−xcos⁡x+sin⁡x]π/3π/2=1−32+π6\begin{aligned} \therefore \mathrm{I} &=\int_{\pi / 3}^{\pi / 2} \mathrm{x} \sin (\pi-\mathrm{x}) \mathrm{dx}=\int_{\pi / 3}^{\pi / 2} \mathrm{x} \sin \mathrm{x} \mathrm{d} \mathrm{x} \\ &=[-\mathrm{x} \cos \mathrm{x}+\sin \mathrm{x}]_{\pi / 3}^{\pi / 2}=1-\frac{\sqrt{3}}{2}+\frac{\pi}{6} \end{aligned}$$
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