Question

Let $f\left(x\right)=x+\varphi \left(x\right)$ where $\varphi \left(x\right)$ is an even function then find the value of ${\int }_{-1}^{1}xf\left(x\right)dx$.

Answer 1

Given,

$f\left(x\right)=x+\varphi \left(x\right)$.

Now $xf\left(x\right)=x^2+x\varphi \left(x\right)$.

Since $\varphi \left(x\right)$ is even then $x\varphi \left(x\right)$ is odd as multiplication of an odd and an even function gives an odd function.

Now,

${\int }_{-1}^{1}xf\left(x\right)dx$

$={\int }_{-1}^1{x}^{2}dx+$${\int }_{-1}^{1}x\varphi \left(x\right)dx$

$={\left[\frac{x^3}{3}\right]}_{x=-1}^{x=1}+0$

$=\frac{2}{3}$.