Q. The value of$\int_{a}^{b}{\frac{x}{|x|}dx,\ }a < b < 0$ is
A) $- (|a| + |d|)$
B) $|b| - |a|$
C) $|a| - |b|$
D) $|a| + |b|$
Explanation
(b)
Let $I = \int_{a}^{b}{\frac{x}{|x|}dx = \int_{a}^{b}{- 1dx}}$
$\Rightarrow \ I = - \lbrack x\rbrack_{a}^{b} = \lbrack - b + a\rbrack$
$= |b| - |a|\ \ \ \ \ \lbrack\because a < b < 0\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.