Chapter 5, Problem 10RQ

To determine

To find: Whether the integral relation ${\displaystyle {\int }_{-5}^5\left(a{x}^2+bx+c\right)dx}=2{\displaystyle {\int }_0^5\left(a{x}^2+c\right)dx}$ is true or false.

Answer to Problem 10RQ

The integral relation ${\displaystyle {\int }_{-5}^5\left(a{x}^2+bx+c\right)dx}=2{\displaystyle {\int }_0^5\left(a{x}^2+c\right)dx}$ is true.

Explanation of Solution

Given:

The integral relation is ${\displaystyle {\int }_{-5}^5\left(a{x}^2+bx+c\right)dx}=2{\displaystyle {\int }_0^5\left(a{x}^2+c\right)dx}$ (1)

The region lies between $x=-5$ and $x=5$.

Theorem used:

1) If $f\left(-x\right)=f\left(x\right)$, that is the equation of the curve is unchanged when x is replaced by $-x$, then the function is an even function and the curve is symmetric about y-axis.

2) If $f\left(-x\right)=-f\left(x\right)$, that is the equation of the curve is changed when x is replaced by $-x$, then the function is an odd function and the curve is symmetric about the origin.

Calculation:

Rearrange the left hand side part of equation (1) as shown below,

${\displaystyle {\int }_{-5}^5\left(a{x}^2+bx+c\right)dx}={\displaystyle {\int }_{-5}^5\left(a{x}^2+c\right)dx}+{\displaystyle {\int }_{-5}^5\left(bx\right)dx}$ (2)

Consider the integral ${\displaystyle {\int }_{-5}^5\left(bx\right)dx}$ (3)

The function $f\left(x\right)$ from equation (3) is,

$f\left(x\right)=bx$ (4)

Substitute $-x$ for x in equation (4), to check the function is odd or even as shown below,

$\begin{array}{c}f\left(-x\right)=b\left(-x\right)\\ =-bx\\ =-f\left(x\right)\end{array}$ (5)

Thus, the function is an odd function at $x=-x$.

From the theorem used, as the function is an odd function, so the value of the function ${\displaystyle {\int }_{-5}^5\left(bx\right)dx}$ is 0.

Substitute $2{\displaystyle {\int }_0^5\left(a{x}^2+c\right)dx}$ for ${\displaystyle {\int }_{-5}^5\left(a{x}^2+c\right)dx}$ and 0 for ${\displaystyle {\int }_{-5}^5\left(bx\right)dx}$ in Equation (2).

$\begin{array}{c}{\displaystyle {\int }_{-5}^5\left(a{x}^2+bx+c\right)dx}=2{\displaystyle {\int }_0^5\left(a{x}^2+c\right)dx}+0\\ =2{\displaystyle {\int }_0^5\left(a{x}^2+c\right)dx}\end{array}$

Thus, the left hand side part is equal to the right hand side part.

Therefore, the statement is true.