Q. The value of integral $\int_{- 1}^{1}\frac{|x + 2|}{x + 2}\ dx$ is
A) 1
B) 2
C) 0
D) $- 1$
Explanation
(b)
Let $I = \int_{- 1}^{1}\frac{|x + 2|}{x + 2}dx$
For $- 1 \leq x \leq 1,|x + 2| = 2 + x$
$\therefore I = \int_{- 1}^{1}\frac{x + 2}{x + 2}\ dx = \int_{- 1}^{1}1dx$
$= \lbrack x\rbrack_{- 1}^{1} = 1 - ( - 1)$
$= 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.