Question

Evalaute the integral
${\int }_{-1}^1\frac{\sin x-x^2}{3-|x|}dx$

• A. $0$
• B. ${\int }_0^1\frac{\sin x}{3-|x|}dx$
• C. ${\int }_0^1\frac{-2x^2}{3-|x|}dx$
• D. $2{\int }_0^1\frac{\sin x-X^2}{3-|x|}dx$

Answer 1

Answer: (C)

${\int }_0^1\frac{-2x^2}{3-|x|}dx$

$I={\int }_{-1}^1\frac{sinx-x^2}{3-\left|x\right|}dx$

$I={\int }_{-1}^1\frac{sin(-x)-x^2}{3-\left|x\right|}dx$

=${\int }_{-1}^1\frac{-sinx-x^2}{3-\left|x\right|}dx$

$2I={\int }_{-1}^1\frac{-2x^2}{3-\left|x\right|}dx$

$I={\int }_{-1}^1\frac{-x^2}{3-\left|x\right|}$

$\left|x\right|,x^2$ symmetrical about

$\frac{1-1}{2}=0$

$\Rightarrow 2{\int }_{-1}^1\frac{-x^2}{3-\left|x\right|}$

$={\int }_0^1\frac{-2x^2}{3-\left|x\right|}dx$