First slide

Evaluation of definite integrals

Question

$\int_{0}^{4} | x - 1 | dx$ is equal to

Moderate

A

$\frac{5}{2}$
B

$\frac{3}{2}$
C

$\frac{1}{2}$
D

5

Solution

Let $\int_{0}^{4} | x - 1 | dx$
It can be seen that, $(x - 1) \leq 0$ when $0 \leq x \leq 1$ and
$(x - 1) \geq 0 when 1 \leq x \leq 4$
$\begin{array}{l} ∴ I = \int_{0}^{1} | x - 1 | dx + \int_{1}^{4} | x - 1 | dx \\ = \int_{0}^{1} (1 - x) dx + \int_{a}^{b} f (x) dx = \int_{a}^{c} f (x) dx + \int_{c}^{b} f (x) dx] \\ = {[x - \frac{x^{2}}{2}]}_{0}^{1} + {[\frac{x^{2}}{2} - x]}_{1}^{4} \\ = (1 - \frac{1}{2}) - 0 + (\frac{4^{2}}{2} - 4) - (\frac{1}{2} - 1) = \frac{1}{2} + 4 + \frac{1}{2} = 5 \end{array}$

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