KCET · Maths · Definite Integration
The value of \(\int_{-1}^{2} \frac{|x|}{x} d x\), is
- A 0
- B 1
- C 2
- D 3
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
Let \(I=\int_{-1}^{2} \frac{|x|}{x} d x=\int_{-1}^{0}\left(-\frac{x}{x}\right) d x+\int_{0}^{2}\left(\frac{x}{x}\right) d x\)
\(\left(\because|x|=\left\{\begin{array}{rll}
x, & \text { if } & x \geq 0 \\
-x, & \text { if } & x < 0
\end{array}\right)\right.\)
\(=-\int_{-1}^{0} d x+\int_{0}^{2} d x\)
\(=-[x]_{-1}^{0}+[x]_{0}^{2}\)
\(=-[0+1]+[2-0]\)
\(=-1+2=1\)
\(\left(\because|x|=\left\{\begin{array}{rll}
x, & \text { if } & x \geq 0 \\
-x, & \text { if } & x < 0
\end{array}\right)\right.\)
\(=-\int_{-1}^{0} d x+\int_{0}^{2} d x\)
\(=-[x]_{-1}^{0}+[x]_{0}^{2}\)
\(=-[0+1]+[2-0]\)
\(=-1+2=1\)
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