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JAMB Mathematics 2016 Paper

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Question : 29

Total: 50

Evaluate

\int\limits_{0}^{\pi/2}

sin xdx

2
-1
1
-2

Solution:

Concept:

Integrate

\sin x

and evaluate it at the given limits.

Explanation:

The antiderivative of

\sin x

is

-\cos x

.

So,

\int_{0}^{\pi/2} \sin x \, dx = -\cos x \Big|_{0}^{\pi/2}

.

Substitute the upper and lower limits:

-\cos\left(\frac{\pi}{2}\right) - \big(-\cos 0\big)

.

Since

\cos\left(\frac{\pi}{2}\right) = 0

and

\cos 0 = 1

, the value is:

-0 - (-1) = 1

.

Answer:

1

, which is option C.

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