Chapter 6.3, Problem 4E

(a)

To determine

To evaluate: The given integral

Answer to Problem 4E

The value of the given integral is $-\sqrt{2}$

Explanation of Solution

Given Information:

The integral ${\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt\text{ =}\sqrt{2}}$ and the integral to be evaluated ${\displaystyle \underset{0}{\overset{-3}{\int }}g(t)dt}$

Formula Used The reverse limit property of a given integral with limits [a, b] can be written as,

  ${\displaystyle \underset{a}{\overset{b}{\int }}g(t)dt\text{ =}}-{\displaystyle \underset{b}{\overset{a}{\int }}g(t)dt}$

Calculation Considering the given integral, ${\displaystyle \underset{0}{\overset{-3}{\int }}g(t)dt}$ , which can be rewritten based on the reverse interval property as,

  ${\displaystyle \underset{0}{\overset{-3}{\int }}g(t)dt}=-{\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt}$

The value of ${\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt}$ is given as $\sqrt{2}$ . Substituting this value in the above expression,

  ${\displaystyle \underset{0}{\overset{-3}{\int }}g(t)dt}=-\sqrt{2}$

(b)

To determine

To evaluate: The given integral

Answer to Problem 4E

The value of the given integral is $\sqrt{2}$

Explanation of Solution

Given Information:

The integral ${\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt\text{ =}\sqrt{2}}$ and the integral ${\displaystyle \underset{-3}{\overset{0}{\int }}g(u)du}$ to be evaluated.

Formula Used For any function g(t) with a finite integration value,

  ${\displaystyle \underset{a}{\overset{b}{\int }}g(t)dt\text{ =}}{\displaystyle \underset{a}{\overset{b}{\int }}g(u)du}$

Here, the variable u is a dummy variable and the value of the integral remains the same irrespective of the variable name.

Calculation Considering the given integral, ${\displaystyle \underset{-3}{\overset{0}{\int }}g(u)du}$ , which can be rewritten based on above mentioned property as,

  ${\displaystyle \underset{-3}{\overset{0}{\int }}g(u)du}={\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt}$

The value of ${\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt}$ is given as $\sqrt{2}$ . Substituting this value in the above expression,

  ${\displaystyle \underset{-3}{\overset{0}{\int }}g(u)du}=\sqrt{2}$

(c)

To determine

To evaluate: The given integral

Answer to Problem 4E

The value of the given integral is $-\sqrt{2}$

Explanation of Solution

Given Information:

The integral ${\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt\text{ =}\sqrt{2}}$ and the integral ${\displaystyle \underset{-3}{\overset{0}{\int }}\left(-g(x)\right)dx}$ to be evaluated.

Formula Used For any function g(t) with a finite integration value,

  ${\displaystyle \underset{a}{\overset{b}{\int }}kg(t)dt\text{ =}k}{\displaystyle \underset{a}{\overset{b}{\int }}g(t)dt}$

Also,

  ${\displaystyle \underset{a}{\overset{b}{\int }}g(t)dt\text{ =}}{\displaystyle \underset{a}{\overset{b}{\int }}g(u)du}$

Here, the variable u is a dummy variable and the value of the integral remains the same irrespective of the variable name.

Calculation Considering the given integral, ${\displaystyle \underset{-3}{\overset{0}{\int }}\left(-g(x)\right)dx}$ , which can be rewritten based on above mentioned properties as,

  ${\displaystyle \underset{-3}{\overset{0}{\int }}\left(-g(x)\right)dx}=-{\displaystyle \underset{-3}{\overset{0}{\int }}g(x)dx}$

  $-{\displaystyle \underset{-3}{\overset{0}{\int }}g(x)dx}=-{\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt}$

The value of ${\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt}$ is given as $\sqrt{2}$ . Substituting this value in the above expression,

  $-{\displaystyle \underset{-3}{\overset{0}{\int }}g(x)dx}=-\sqrt{2}$

(d)

To determine

To evaluate: The given integral

Answer to Problem 4E

The value of the given integral is $1$

Explanation of Solution

Given Information:

The integral ${\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt\text{ =}\sqrt{2}}$ and the integral ${\displaystyle \underset{-3}{\overset{0}{\int }}\frac{g(r)}{\sqrt{2}}dr}$ is to be evaluated.

Formula Used:

For any function g(t) with a finite integration value,

  ${\displaystyle \underset{a}{\overset{b}{\int }}kg(t)dt\text{ =}k}{\displaystyle \underset{a}{\overset{b}{\int }}g(t)dt}$

Also,

  ${\displaystyle \underset{a}{\overset{b}{\int }}g(t)dt\text{ =}}{\displaystyle \underset{a}{\overset{b}{\int }}g(r)dr}$

Here, the variable r is a dummy variable and the value of the integral remains the same irrespective of the variable name.

Calculation Considering the given integral, ${\displaystyle \underset{-3}{\overset{0}{\int }}\frac{g(r)}{\sqrt{2}}dr}$ , which can be rewritten based on above mentioned properties as,

  ${\displaystyle \underset{-3}{\overset{0}{\int }}\frac{g(r)}{\sqrt{2}}dr}=\frac{1}{\sqrt{2}}{\displaystyle \underset{-3}{\overset{0}{\int }}g(r)dr}$

  $\frac{1}{\sqrt{2}}{\displaystyle \underset{-3}{\overset{0}{\int }}g(r)dr}=\frac{1}{\sqrt{2}}{\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt}$

The value of ${\displaystyle \underset{-3}{\overset{0}{\int }}g(t)dt}$ is given as $\sqrt{2}$ . Substituting this value in the above expression,

  $\begin{array}{c}\frac{1}{\sqrt{2}}{\displaystyle \underset{-3}{\overset{0}{\int }}g(r)dr}=\frac{\sqrt{2}}{\sqrt{2}}\\ =1\end{array}$