Chapter 5, Problem 5RQ

To determine

To find: Whether the given statement is true or false.

Answer to Problem 5RQ

The given statement is false.

Explanation of Solution

Given:

The statement is that “If f is continuous $\left[a,b\right]$ and $f\left(x\right)\ge 0$ then ${\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{f\left(x\right)}}dx=\sqrt{{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx}$ (1)

Calculation:

The statement is proved false by counter example.

Let the function $f\left(x\right)=x^2$, upper limit b as 1 and lower limit a as 0.

Substitute $x^2$ for $f\left(x\right)$, 0 for a, and 1 for b in equation (1).

${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x^2}}dx=\sqrt{{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx}$ (2)

Evaluate the integral ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x^2}}dx$ in equation (2).

$\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x^2}}dx={\displaystyle \underset{0}{\overset{1}{\int }}{\left(x^2\right)}^{\frac{1}{2}}}dx\\ ={\displaystyle \underset{0}{\overset{1}{\int }}x}dx\\ ={\left(\frac{x^2}{2}\right)}_0^1\\ =\frac{1}{2}\end{array}$ (3)

Evaluate the of the integral $\sqrt{{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx}$ in equation (2).

$\begin{array}{c}\sqrt{{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx}=\sqrt{{\left(\frac{x^3}{3}\right)}_0^1}\\ =\sqrt{\frac{1}{3}-\frac{0}{3}}\\ =\sqrt{\frac{1}{3}}\end{array}$ (4)

From equations (3) and (4), it is observed that the values of the integral is not equal that is,

${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x^2}}dx\ne \sqrt{{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2}dx}$

Hence, equation (1) is not true that is, ${\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{f\left(x\right)}}dx\ne \sqrt{{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx}$.

Therefore, the given statement is false.