Q229 / 270 · NDA 2012 (Phase 1) · Mathematics

What is $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\sin x| d x$ equal to?

  • Correct answer
    A
    2
  • B
    1
  • C
    p
  • D
    0

Solution

$\begin{aligned} & \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\sin x| d x \\ & |\sin x|= \begin{cases}\sin x & 0 \lt x \lt \frac{\pi}{2} \\ -\sin x & -\frac{\pi}{2} \lt x \lt 0\end{cases} \\ & \frac{\pi}{2}\end{aligned} \begin{aligned} & =\int_{-\frac{\pi}{2}}^0-\sin x d x+\int_0^{\sin x d x} \\ & =-[-\cos x]_{-\pi / 2}^0+(-\cos x) \int_0^{\pi / 2} \\ & =(\cos 0-\cos (-\pi / 2)+(-\cos \pi / 2-(-\cos 0)) \\ & =(1-0)+(-0+1)=2 .\end{aligned}$

Hence, the correct answer is option: 'A'.

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