Jambmaths question:
Evaluate $\int\limits_{0}^{\tfrac{\pi }{2}}{\sin xdx}$
Option A:
2
Option B:
–1
Option C:
1
Option D:
–2
Jamb Maths Solution:
$\begin{align} & \int\limits_{0}^{\tfrac{\pi }{2}}{\sin xdx}=\left[ -\cos x \right]_{0}^{\tfrac{\pi }{2}}=[-\cos \tfrac{\pi }{2}-(-\cos 0)]= \\ & \int\limits_{0}^{\tfrac{\pi }{2}}{\sin xdx}=0-(-1)=1 \\\end{align}$
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