Chapter 6.4, Problem 76E

To determine

To calculate:the value of ${\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)}dt$ is even

Answer to Problem 76E

  ${\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)}dt$ is even

Explanation of Solution

Given information:

If f is an odd continuous function based on the graphical argument ${\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)}dt$ is even

Formula used:

  ${\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt=-{\displaystyle \underset{-x}{\overset{0}{\int }}f\left(t\right)dt}}$

Calculation:

If a function is odd then the area from 0 to x is opposite to area from -x to 0.

  ${\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt=-{\displaystyle \underset{-x}{\overset{0}{\int }}f\left(t\right)dt}}$

The graph is

  

Now,

  ${\displaystyle \underset{b}{\overset{b}{\int }}f\left(x\right)}dx=-{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$

So,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{-x}{\int }}f\left(t\right)dt=-{\displaystyle \underset{-x}{\overset{0}{\int }}f\left(t\right)dt}}\\ =-\left[-{\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}\right]\\ {\displaystyle \underset{0}{\overset{-x}{\int }}f\left(t\right)dt={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}}\end{array}$

Since for a function to be even,

  $f\left(-x\right)=f\left(x\right)$

Hence, ${\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)}dt$ is even

Conclusion:

Hence, ${\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)}dt$ is even