Evaluate $\displaystyle\int_{0}^{\pi/2} \sin x\,dx$.

Question

Accepted Answer

B. $1$

$\displaystyle\int_{0}^{\pi/2} \sin x\,dx = [-\cos x]_{0}^{\pi/2} = -\cos(\pi/2) + \cos(0) = -0 + 1 = 1$.
Geometric interpretation: this is the area under one quarter of the sine curve from $0$ to $\pi/2$, which equals exactly $1$ in standard units.

Answer

$-1$

Answer

$\pi/2$

Answer

$0$

Evaluate $\displaystyle\int_{0}^{\pi/2} \sin x\,dx$.

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