Chapter 9.4, Problem 2QR

To determine

To find: the value of integral.

Answer to Problem 2QR

The value of integral is $0$ .

Explanation of Solution

Given information:

Theintegral is ${\displaystyle {\int }_{-1}^1\frac{x}{x^2+1}dx}$ .

Calculation: to evaluate the integral, first find that whether the function is even or odd,

  $\begin{array}{l}{\displaystyle {\int }_{-1}^1\frac{x}{x^2+1}dx}={\displaystyle {\int }_{-1}^{1}f\left(x\right)dx}\\ f\left(x\right)=\frac{x}{x^2+1}\\ f\left(-x\right)=\frac{-x}{{\left(-x\right)}^2+1}\text{ [substitute }x=-x]\\ f\left(-x\right)=-\frac{x}{x^2+1}\\ f\left(-x\right)=-f\left(x\right)\end{array}$

Because the function is odd, so the property of definite function,

  ${\displaystyle {\int }_{-1}^1\frac{x}{x^2+1}dx}=0\text{ [}{\displaystyle {\int }_{-a}^{a}f\left(x\right)dx=0\text{ when }f\left(-x\right)=f\left(x\right)]}$

Thus, the value is $0$ .