Chapter 7, Problem 4RE

To determine

To evaluate the integral.

Answer to Problem 4RE

The required solution is , ${\displaystyle {\int }_{-1}^{1}2x\sin \left(1-x^2\right)dx}=0$ .

Explanation of Solution

Given information: The equation is, ${\displaystyle {\int }_{-1}^{1}2x\sin \left(1-x^2\right)dx}$ .

Concept used: Definite integrals,

The definite integral of $f\left(x\right)$ over $\left[a,b\right]$ , denoted by ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx,}$ is given by, ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx,}={\left[F\left(\text{b}\right)\right]}_a^b=F\left(b\right)-F\left(a\right)$

Calculation : The required solution is obtained as,

  $\begin{array}{l}{\displaystyle {\int }_{-1}^{1}2x\sin \left(1-x^2\right)dx}\\ \text{ let }u=\left(1-x^2\right)u=1-{\left(-1\right)}^2=0\\ du=-2xdxu=1-{\left(1\right)}^2=0\\ -du=2xdx\\ ={\displaystyle {\int }_0^0\sin u}\left(-du\right)\\ =-{\displaystyle {\int }_0^0\sin u}du\\ =0\left(\text{since the bound of the integral are same, the value is 0}\right)\end{array}$

Hence, the required solution is ${\displaystyle {\int }_{-1}^{1}2x\sin \left(1-x^2\right)dx}=0$ .