Question 4

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Jambmaths question: 

Evaluate $\int_{0}^{\tfrac{\pi }{2}}{\sin 2xdx}$

Option A: 

– ½

Option B: 

1

Option C: 

–1

Option D: 

0

Jamb Maths Solution: 

$\begin{align}  & \int_{0}^{\tfrac{\pi }{2}}{\sin 2xdx}=\left[ -\tfrac{1}{2}\cos 2x \right]_{0}^{\tfrac{\pi }{2}}=-\tfrac{1}{2}\left[ \cos 2x \right]_{0}^{\tfrac{\pi }{2}} \\ & =-\tfrac{1}{2}\left[ \cos 2(\tfrac{\pi }{2})-\cos 2(0) \right] \\ & =-\tfrac{1}{2}\left[ \cos \pi -\cos 0 \right]=-\tfrac{1}{2}\left[ -1-1 \right]=1 \\\end{align}$

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