Question

The value of integral $\underset{-1}{\overset{1}{\int }}\frac{|x+2|}{x+2}dx$ is

• A. $1$
• B. $2$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

Let $I=\underset{-1}{\overset{1}{\int }}\frac{|x+2|}{x+2}dx$

For $-1\le x\le 1,\left|x+2\right|=2+x$

$\therefore I=\underset{-1}{\overset{1}{\int }}\frac{x+2}{x+2}dx$

$=\underset{-1}{\overset{1}{\int }}1dx$

$={\left[x\right]}_{-1}^1=1-\left(-1\right)$

$=2$