First slide

Evaluation of definite integrals

Question

$\int_{0}^{π / 2} min (\sin x, \cos x) d x$ equals to

Moderate

A

$2 \sqrt{2}$
B

$\sqrt{2}$
C

$2 - \sqrt{2}$
D

$2 + \sqrt{2}$

Solution

$\begin{array}{l} \int_{0}^{π / 2} min (\sin x, \cos x) d x \\ = \int_{0}^{π / 4} min (\sin x, \cos x) d x + \int_{π / 4}^{π / 2} min (\sin x, \cos x) d x \\ = \int_{0}^{π / 4} \sin x d x + \int_{π / 4}^{π / 2} \cos x d x \\ = - {\cos x|}_{0}^{π / 4} + {\sin x|}_{π / 4}^{π / 2} \\ = - [\frac{1}{\sqrt{2}} - 1] + [1 - \frac{1}{\sqrt{2}}] = 2 - \sqrt{2} \end{array}$

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