$\int_{-1}^1 \frac{|x|}{x} d x$ is equal to
$\int_{-1}^1 \frac{|x|}{x} d x$ is equal to
- 1
- -1
- 0
- $\frac{1}{2}$
Solution
$
\begin{aligned}
\text { } \int_{-1}^1 \frac{|x|}{x} d x & =1 \\
f(x) & =\frac{|x|}{x}=\left\{\begin{array}{cc}
-\frac{x}{x}, & x < 0 \\
\frac{x}{x}, & x>0
\end{array}=\left\{\begin{array}{cc}
-1, & x < 0 \\
1, & x>0
\end{array}\right.\right. \\
\therefore \quad I & =\int_{-1}^0(-1) d x+\int_0^1(1) d x \\
& =[-x]_{-1}^0+[x]_0^1=-1+1=0
\end{aligned}
$
Asked in: AP EAMCET 2021 (23 Aug Shift 1)
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