JAMB 2018 · UTME · Q34

Evaluate ∫−π/2π/2cos⁡x dx\displaystyle\int_{-\pi/2}^{\pi/2} \cos x\,dx.

Worked solution (try it first)
  1. cos⁡x\cos x integrates to sin⁡x\sin x.
  2. [sin⁡x]−π/2π/2=sin⁡π2−sin⁡(−π2)\left[\sin x\right]_{-\pi/2}^{\pi/2} = \sin\frac\pi2 - \sin\left(-\frac\pi2\right)
    =1−(−1)= 1 - (-1)
    =2= 2, option C.

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