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UPSC NDA I 2016 Mathematics Solved Paper

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Question : 86 of 120

Marks: +1, -0

What is

\overset{4\pi}{\underset{0}{\int\limits}} |\cos x|\,dx

equal to?

0
2
4
8

Solution:

Concept:

The modulus function

|\cos x|

is periodic with period

\pi

.

Using the property

\int\limits_{0}^{nT} f(x)\,dx = n \int\limits_{0}^{T} f(x)\,dx

for a periodic function with period

T

.

Explanation:

Let

I = \overset{4\pi}{\underset{0}{\int\limits}} |\cos x|\,dx

.

Since the period of

|\cos x|

is

\pi

, we have

I = 4 \overset{\pi}{\underset{0}{\int\limits}} |\cos x|\,dx

.

On the interval

[0, \pi]

,

|\cos x|

is defined piecewise:

• For

0 \le x < \frac{\pi}{2}

,

\cos x \ge 0

so

|\cos x| = \cos x

.

• For

\frac{\pi}{2} \le x < \pi

,

\cos x \le 0

so

|\cos x| = -\cos x

.

Thus,

I = 4\left[ \overset{\frac{\pi}{2}}{\underset{0}{\int\limits}} \cos x\,dx + \overset{\pi}{\underset{\frac{\pi}{2}}{\int\limits}} (-\cos x)\,dx \right]

.

Integrating:

\int \cos x\,dx = \sin x

, and

\int (-\cos x)\,dx = -\sin x

.

So

I = 4[\sin x]_{0}^{\frac{\pi}{2}} - 4[\sin x]_{\frac{\pi}{2}}^{\pi}

.

Evaluating:

[\sin x]_{0}^{\frac{\pi}{2}} = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = 1

.

[\sin x]_{\frac{\pi}{2}}^{\pi} = \sin\pi - \sin\frac{\pi}{2} = 0 - 1 = -1

.

Thus,

I = 4 \times 1 - 4 \times (-1) = 4 + 4 = 8

.

Answer:

8

(Option D).

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