First slide

Evaluation of definite integrals

Question

$\int_{0}^{2} | \sin π x | d x =$

Moderate

A

$\frac{2}{π}$
B

$\frac{1}{π}$
C

$\frac{3}{π}$
D

$\frac{4}{π}$

Solution

$\int_{0}^{2} | \sin π x | d x = \int_{0}^{1} | \sin π x | d x + \int_{1}^{2} | \sin π x | d x$
$= \int_{0}^{1} \sin (π x) d x + \int_{1}^{2} (- \sin π x) d x$

$= {[- \frac{\cos π x}{x}]}_{0}^{1} + {[\frac{\cos π x}{π}]}_{1}^{2}$

$= - \frac{1}{π} [- 1 - 1] + \frac{1}{π} [1 - (- 1)]$ $= \frac{2}{π} + \frac{2}{π} = \frac{4}{π}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App