₀^{π/2} | x - x|\,dx is equal to

Correct Answer is : 2(2−1)2≤ft( { 2 - 1} )2(2−1)

Correct Answer is : 2(2−1)2≤ft( { 2 - 1} )2(2−1)

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UPSC NDA I 2019 Mathematics Solved Paper

Question : 82 of 120

Marks: +1, -0

\int_{0}^{\frac{\pi}{2}} |\sin x - \cos x|\,dx

is equal to

Solution:

Concept:

The integrand

|\sin x - \cos x|

changes its behaviour at

x = \frac{\pi}{4}

, where

\sin x = \cos x

Explanation:

For

0 < x < \frac{\pi}{4}

\sin x < \cos x

, so

\sin x - \cos x < 0

, and

|\sin x - \cos x| = \cos x - \sin x

For

\frac{\pi}{4} < x < \frac{\pi}{2}

\sin x > \cos x

, so

\sin x - \cos x > 0

, and

|\sin x - \cos x| = \sin x - \cos x

Thus the integral splits as:

\int_0^{\frac{\pi}{2}} |\sin x - \cos x| \, dx = \int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx

Evaluate the first part:

\int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx = \left[ \sin x + \cos x \right]_0^{\frac{\pi}{4}} = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - (0 + 1) = \sqrt{2} - 1

Evaluate the second part:

\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx = \left[ -\cos x - \sin x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \left( -\cos\frac{\pi}{2} - \sin\frac{\pi}{2} \right) - \left( -\cos\frac{\pi}{4} - \sin\frac{\pi}{4} \right) = (0 - 1) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) = -1 + \sqrt{2} = \sqrt{2} - 1

Adding both parts:

(\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)

Answer:

2(\sqrt{2} - 1)

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