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NDA 2012
Exam
NDA
Year
2012
Subject
Mathematics
Chapter
Definite Integration

NDA 2012 · Mathematics · Definite Integration

What is ∫_ -(π)/(2) ^ (π)/(2) | x| d x equal to?

What is ∫−π2π2∣sin⁡x∣dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\sin x| d x equal to?

Options

  1. A2
  2. B1
  3. Cp
  4. D0

Correct answer

A. 2

Step-by-step solution

∫−π2π2∣sin⁡x∣dx∣sin⁡x∣={sin⁡x0<x<π2−sin⁡x−π2<x<0π2\begin{aligned} & \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\sin x| d x \\ & |\sin x|= \begin{cases}\sin x & 0 < x < \frac{\pi}{2} \\ -\sin x & -\frac{\pi}{2} < x < 0\end{cases} \\ & \frac{\pi}{2}\end{aligned}=∫−π20−sin⁡xdx+∫0sin⁡xdx=−[−cos⁡x]−π/20+(−cos⁡x)∫0π/2=(cos⁡0−cos⁡(−π/2)+(−cos⁡π/2−(−cos⁡0))=(1−0)+(−0+1)=2.\begin{aligned} & =\int_{-\frac{\pi}{2}}^0-\sin x d x+\int_0^{\sin x d x} \\ & =-[-\cos x]_{-\pi / 2}^0+(-\cos x) \int_0^{\pi / 2} \\ & =(\cos 0-\cos (-\pi / 2)+(-\cos \pi / 2-(-\cos 0)) \\ & =(1-0)+(-0+1)=2.\end{aligned}$
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