Chapter 5, Problem 1RE

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume f and f′ are continuous functions for all real numbers.

1. a. If $A(x)={\displaystyle {\int }_a^{x}f(t)dt}$, and f(t) = 2t − 3, then A is a quadratic function.
2. b. Given an area function $A(x)={\displaystyle {\int }_a^{x}f(t)dt}$ and an antiderivative F of f, it follows that A′(x) = F(x).
3. c. ${\displaystyle {\int }_a^b{f}^{\prime }(x)dx=f(b)-f(a)}$.
4. d. If f is continuous on [a, b] and ${\displaystyle {\int }_a^b|f(x)|dx=0}$, then f(x) = 0 on [a, b].
5. e. If the average value of f on [a, b] is zero, then f(x) = 0 on [a, b].
6. f. ${\displaystyle {\int }_a^b(2f(x)-3g(x))dx=2{\displaystyle {\int }_a^{b}f(x)dx+3{\displaystyle {\int }_b^{a}g(x)dx}}}$.
7. g. ${\displaystyle \int f^{\prime }(g(x))g^{\prime }(x)dx=f(g(x))+C}$.

To determine

Whether the statement, “If $A\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}$ and $f\left(t\right)=2t-3$, then A is quadratic function” is true or false.

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Given that the functions f and f’ are continuous for all real numbers.

Consider $A\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}$ where $f\left(t\right)=2t-3$.

Evaluate ${\displaystyle \underset{a}{\overset{x}{\int }}\left(2t-3\right)dt}$.

$\begin{array}{c}A\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}\left(2t-3\right)dt}\\ ={\displaystyle \underset{a}{\overset{x}{\int }}2tdt}-{\displaystyle \underset{a}{\overset{x}{\int }}3dt}\\ =2{\left[\frac{t^2}{2}\right]}_a^x-3{\left[t\right]}_a^x\\ =x^2-3x-a^2+3a\end{array}$

Therefore, $A\left(x\right)=x^2-3x-a^2+3a$ is a quadratic function.

Thus, the given statement is true.

To determine

Whether the statement, “Given an area function $A\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}$ and an antiderivative F of f, it follows that $A^{\prime }\left(x\right)=F\left(x\right)$” is true or false.

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Give that the functions f and f’ are continuous for all real numbers.

Let $A\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$ and an antiderivative F of f be $F\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$.

Differentiate $A\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}$ with respect to x.

$\begin{array}{c}{A}^{\prime }\left(x\right)=\frac{d}{dx}{\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}\\ =f\left(x\right)\end{array}$

Here $A^{\prime }\left(x\right)=f\left(x\right)$ but not equal to $F\left(x\right)$.

Thus, the given statement is false.

To determine

Whether the statement, “${\displaystyle \underset{a}{\overset{b}{\int }}{f}^{\prime }\left(x\right)dx=}f\left(b\right)-f\left(a\right)$” is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Evaluate ${\displaystyle \underset{a}{\overset{b}{\int }}{f}^{\prime }\left(x\right)dx}$.

$\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}{f}^{\prime }\left(x\right)dx}=\frac{d}{dx}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}\\ ={\left[f\left(x\right)\right]}_a^b\\ =f\left(b\right)-f\left(a\right)\end{array}$

Therefore, the statement is true.

To determine

Whether the statement, “If f is continuous on $\left[a,b\right]$ and ${\displaystyle \underset{a}{\overset{b}{\int }}\left|f\left(x\right)\right|dx}$, then $f\left(x\right)=0$ on $\left[a,b\right]$” is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Recall the known fact that $\left|f\left(x\right)\right|$ is always positive and therefore, the integral function ${\displaystyle \underset{a}{\overset{b}{\int }}\left|f\left(x\right)\right|dx}$ is also positive.

Note that if f is continuous on $\left[a,b\right]$, then $f\left(c\right)=0$ at some constant c.

Given that f is continuous on $\left[a,b\right]$ and ${\displaystyle \underset{a}{\overset{b}{\int }}\left|f\left(x\right)\right|dx}$ is positive, and this implies that constantly $f\left(x\right)=0$.

Therefore, the statement is true.

To determine

Whether the statement, “If the average value of f on $\left[a,b\right]$ is zero, then $f\left(x\right)=0$ on $\left[a,b\right]$” is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The following example will disprove the given statement.

Consider the trigonometric function $f\left(x\right)=\sin x$ over the interval $\left[0,2\pi \right]$.

The formula for the average function value is $f_{average}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$.

Evaluate $f_{average}=\frac{1}{2\pi -0}{\displaystyle \underset{0}{\overset{2\pi }{\int }}\sin xdx}$.

$\begin{array}{c}{f}_{average}=\frac{1}{2\pi -0}{\displaystyle \underset{0}{\overset{2\pi }{\int }}\sin xdx}\\ =\frac{1}{2\pi }{\left[-\cos x\right]}_0^{2\pi }\\ =\frac{1}{2\pi }\left[-\cos 2\pi -\left(-\cos \left(0\right)\right)\right]\\ =\frac{1}{2\pi }\left(-1+1\right)\end{array}$

It can be simplified as $f_{average}=0$.

From the above counterexample, it is proved that the statement is false.

Therefore, the statement is false.

To determine

Whether the statement, “${\displaystyle \underset{a}{\overset{b}{\int }}\left(2f\left(x\right)-3g\left(x\right)\right)dx=}2{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}+3{\displaystyle \underset{b}{\overset{a}{\int }}g\left(x\right)dx}$”, is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Evaluate ${\displaystyle \underset{a}{\overset{b}{\int }}\left(2f\left(x\right)-3g\left(x\right)\right)dx}$.

$\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}\left(2f\left(x\right)-3g\left(x\right)\right)dx}={\displaystyle \underset{a}{\overset{b}{\int }}2f\left(x\right)dx}-{\displaystyle \underset{a}{\overset{b}{\int }}2f\left(x\right)dx}\\ =2{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}-{\displaystyle \underset{a}{\overset{b}{\int }}3f\left(x\right)dx}\left(\because {\displaystyle \underset{a}{\overset{b}{\int }}cf\left(x\right)dx=c{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}}\right)\\ =2{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}-\left(-{\displaystyle \underset{b}{\overset{a}{\int }}3f\left(x\right)dx}\right)\left(\because {\displaystyle \underset{b}{\overset{a}{\int }}f\left(x\right)dx=-{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}}\right)\\ =2{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}+{\displaystyle \underset{b}{\overset{a}{\int }}3f\left(x\right)dx}\end{array}$

Therefore, the statement is true.

To determine

Whether the statement, “${\displaystyle \int f^{\prime }\left(g\left(x\right)\right)g^{\prime }\left(x\right)dx=f\left(g\left(x\right)\right)}+C$” is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Consider ${\displaystyle \int f^{\prime }\left(g\left(x\right)\right)g^{\prime }\left(x\right)dx}$.

Consider the change of variable function as $u=g\left(x\right)$.

Then, $du=g^{\prime }\left(x\right)dx$.

Therefore, the given integral becomes,

$\begin{array}{c}{\displaystyle \int f^{\prime }\left(g\left(x\right)\right)g^{\prime }\left(x\right)dx}={\displaystyle \int f^{\prime }\left(u\right)du}\\ =f\left(u\right)+C\\ =f\left(g\left(x\right)\right)+C\left(\because u=g\left(x\right)\right)\end{array}$

Therefore, the statement is true.