Chapter 6, Problem 1CRQ

(a)

To determine

To fill: A function F is an anti-derivative of the function f on an interval $I$ if ______ for all $x$ in $I$.

Answer to Problem 1CRQ

A function F is an anti-derivative of the function f on an interval $I$ if $F^{\prime }\left(x\right)=f\left(x\right)$ for all $x$ in $I$.

Explanation of Solution

Let $f\left(x\right)$ be a continuous function on a closed interval $\left[a,b\right]$. Then, the anti-derivative for the given function is $F\left(x\right)$ if an only if $F\left(x\right)$ is a continuous function on a closed interval $\left[a,b\right]$ and $F^{\prime }\left(x\right)=f\left(x\right)$ for all $x$ in $\left[a,b\right]$. It is represents as,

$F^{\prime }\left(x\right)=f\left(x\right)$ (1)

Differentiate equation (1) on both sides with respect to x.

$\begin{array}{c}\frac{d}{dx}\left({\displaystyle \int f\left(x\right)}dx\right)=\frac{d}{dx}\left(F\left(x\right)+C\right)\\ f\left(x\right)=\frac{d}{dx}\left(F\left(x\right)\right)+\frac{d}{dx}\left(C\right)\\ =F^{\prime }\left(x\right)+0\\ =F^{\prime }\left(x\right)\end{array}$

Let a function $f\left(x\right)=2x$.

$\begin{array}{c}{\displaystyle \int f\left(x\right)}dx={\displaystyle \int 2x}dx\\ =2\left(\frac{x^2}{2}\right)+C\\ =x^2+C\end{array}$ (2)

Let $F\left(x\right)={\displaystyle \int f\left(x\right)}$.

Differentiate equation (2) on both sides with respect to $x$ and evaluate $F^{\prime }\left(x\right)$.

$\begin{array}{c}{F}^{\prime }\left(x\right)=\frac{d}{dx}\left(x^2+C\right)\\ =\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(C\right)\\ =2x\\ =f\left(x\right)\end{array}$

The function $F\left(x\right)=x^2$ is anti-derivative of $f\left(x\right)=2x$.

Thus, the function $F$ is an anti-derivative of $f$ on an interval $I$ if $F^{\prime }\left(x\right)=f\left(x\right)$ for all $x$ in $I$.

(b)

To determine

To fill: If $F$ is an anti-derivative of $f$ on an interval $I$, then every anti-derivative of $f$ on $I$ has the form _____.

Answer to Problem 1CRQ

If $F$ is an anti-derivative of $f$ on an interval $I$, then every anti-derivative of $f$ on $I$ has the form $F\left(x\right)+C$.

Explanation of Solution

Let $f\left(x\right)$ be a continuous function on a closed interval $\left[a,b\right]$. Then, the anti-derivative for the given function is $F\left(x\right)$ if an only if $F\left(x\right)$ is a continuous function on a closed interval $\left[a,b\right]$ and $F^{\prime }\left(x\right)=f\left(x\right)$ for all $x$ in $\left[a,b\right]$. It is represents as,

${\displaystyle \int f\left(x\right)}dx=F\left(x\right)+C$.

Let a function $f\left(x\right)=\frac{3}{5}{x}^2$.

$\begin{array}{c}{\displaystyle \int f\left(x\right)}dx={\displaystyle \int \frac{3}{5}{x}^2}dx\\ =\frac{3}{5}\left(\frac{x^3}{3}\right)+C\\ =\frac{x^3}{5}+C\\ =F\left(x\right)+C\end{array}$

The anti-derivative of the function $f\left(x\right)$ is of the form $F\left(x\right)+C$.

Thus, if $F$ is an anti-derivative of $f$ on an interval $I$, then every anti-derivative of $f$ on $I$ has the form $F\left(x\right)+C$.