Question

${\int }_{-1}^1\frac{\sin x-x^2}{3-\left|x\right|}=p{\int }_0^1\frac{-x^2}{3-\left|x\right|}dx$.
Find the value of p

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

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

$=I_1+I_2$

where $I_1$ is an odd function and $I_2$ is even function

$I=I_2=2{\int }_0^1\frac{-x^2}{3-\left|x\right|}dx\Rightarrow p=2$