First slide

Evaluation of definite integrals

Question

$\int_{0}^{2} | 1 - x | d x =$

Easy

A

0
B

1
C

$\frac{1}{2}$
D

$- \frac{1}{2}$

Solution

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

                   $= {(x - \frac{x^{2}}{2})}_{0}^{1} + {(\frac{x^{2}}{2} - x)}_{1}^{2}$

                   $= 1$

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