Question
Evaluate: $\int_{-1}^2 \frac{|x|}{x} d x$
✓
Answer
Let, $I=\int_{-1}^2 \frac{|x|}{x} d x$
Since,
$\begin{aligned} \frac{|x|}{x} & =\left\{\begin{array}{l}\frac{-x}{x}, x<0 \\ \frac{x}{x}, x>0\end{array}\right. \\ & =\left\{\begin{array}{l}-1, x<0 \\ 1, x>0\end{array}\right.\end{aligned}$
$\begin{aligned} \therefore \quad I & =\int_{-1}^0(-1) d x+\int_0^2(1) d x \\ & =[-x]_{-1}^0+[x]_0^2 \\ & =-[0-(-1)]+(2-0) \\ & =-1+2 \\ & =1\end{aligned}$
Since,
$\begin{aligned} \frac{|x|}{x} & =\left\{\begin{array}{l}\frac{-x}{x}, x<0 \\ \frac{x}{x}, x>0\end{array}\right. \\ & =\left\{\begin{array}{l}-1, x<0 \\ 1, x>0\end{array}\right.\end{aligned}$
$\begin{aligned} \therefore \quad I & =\int_{-1}^0(-1) d x+\int_0^2(1) d x \\ & =[-x]_{-1}^0+[x]_0^2 \\ & =-[0-(-1)]+(2-0) \\ & =-1+2 \\ & =1\end{aligned}$
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