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JAMB Mathematics 2013 Paper

Question : 47

Total: 47

Evaluate

\int\limits_{0}^{\frac{\pi}{2}} \sin x \, dx

Solution:

Concept:

The definite integral of

\sin x

is obtained by finding its antiderivative and then applying the limits of integration.

Explanation:

The antiderivative of

\sin x

-\cos x

So, the integral becomes:

\int_{0}^{\frac{\pi}{2}} \sin x \, dx = \left[-\cos x\right]_{0}^{\frac{\pi}{2}}

Now evaluate at the upper and lower limits:

\left[-\cos x\right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - \left(-\cos 0\right)

Since

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

and

-\cos 0 = -1

, we have:

0 - (-1) = 1

Answer:

The value of the integral is

1

. Therefore, the correct option is C.

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