First slide

Evaluation of definite integrals

Question

$The value of \int_{a}^{b} \frac{| x |}{x} d x is$

Moderate

A

$| b | - | a |$
B

$| a | - | b |$
C

$| b | + | a |$
D

$- | b | - | a |$

Solution

$Case I: If 0 \leq a < b, then \frac{| x |}{x} = 1$
$∴ I = \int_{a}^{b} 1 . d x = b - a = | b | - | a |$
$\begin{aligned} Case II: If a < b \leq 0, then | x | = - x \\ \begin{aligned} I & = \int_{a}^{b} \frac{- x}{x} d x = \int_{a}^{b} (- 1) d x \\ = [- x]_{a}^{b} = - b - (- a) = | b | - | a | \end{aligned} \end{aligned}$
$Case III: If a < 0 < b$
$then | x | = - x when a < x < 0$
$and | x | = x when 0 < x < b$
$\begin{array}{l} \begin{aligned} I & = \int_{a}^{b | x |} \frac{| x |}{x} d x = \int_{a}^{0} \frac{| x |}{x} d x + \int_{0}^{b} \frac{| x |}{x} d x \\ = \int_{a}^{0} \frac{- x}{x} d x + \int_{0}^{b} \frac{x}{x} d x \\ = \int_{a}^{0} (- 1) d x + \int_{0}^{b} 1 d x \\ = [- x]_{a}^{0} + [x]_{0}^{b} = a + b = b - (- a) = | b | - | a | \end{aligned} \\ Hence, in all the cases I = \int_{a}^{b} \frac{| x |}{x} d x = | b | - | a | \end{array}$

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