First slide

Evaluation of definite integrals

Question

$\int_{- 5}^{5} | x + 2 | dx$ is equal to

Moderate

A

13
B

19
C

21
D

29

Solution

Here, the given integrand is in the form of absolute function and we define the absolute function as $| x | = x, x \geq 0$
or $| x | = - x, x < 0$ by using it, we convert the given integrand in simple form and then integrate it.
Let, $I = \int_{- 5}^{5} | x + 2 | dx$
It can be seen that $(x + 2) \leq 0 on [- 5, - 2] and (x + 2) \geq 0$ on [-2,5]
$\begin{aligned} \begin{aligned} ∴ I & = \int_{- 5}^{- 2} - (x + 2) dx + \int_{- 2}^{5} (x + 2) dx \\ [∵ \int_{a}^{b} f (x) dx = \int_{a}^{c} f (x) dx + \int_{c}^{b} f (x) dx] \end{aligned} \\ \Rightarrow I = - {[\frac{x^{2}}{2} + 2 x]}_{- 5}^{- 2} + {[\frac{x^{2}}{2} + 2 x]}_{- 2}^{5} \end{aligned}$
$\begin{array}{l} = - [\frac{(- 2)^{2}}{2} + 2 (- 2) - \frac{(- 5)^{2}}{2} - 2 (- 5)] + [\frac{(5)^{2}}{2} + 2 (5) - \frac{(- 2)^{2}}{2} - 2 (- 2)] \\ = - [2 - 4 - \frac{25}{2} + 10] + [\frac{25}{2} + 10 - 2 + 4] \\ = - 2 + 4 + \frac{25}{2} - 10 + \frac{25}{2} + 10 - 2 + 4 = 29 \end{array}$

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