Chapter 4.8, Problem 21E

To determine

To calculate: The function $f\left(x\right)$ from the equation $f^{\text{'}\text{'}}\left(x\right)=\frac{2}{3}{x}^{2/3}$ .

Answer to Problem 21E

The function $f\left(x\right)=\frac{3}{20}{x}^{\frac{8}{3}}+Cx+D$ from the equation $f^{\text{'}\text{'}}\left(x\right)=\frac{2}{3}{x}^{2/3}$ .

Explanation of Solution

Given information:

The equation is given as:

  $f^{\text{'}\text{'}}\left(x\right)=\frac{2}{3}{x}^{2/3}$

Formula used:

The function F is antiderivative of $f\left(x\right)$ on an interval $I$ if $F^{`}\left(x\right)=f\left(x\right)$ for all $x$ in $I$

The function $F$ is known as antiderivative of $f\left(x\right)$ on an interval $I$ the most general antiderivative of the function on interval $I$ is $F\left(x\right)+C$ , where $C$ is an arbitrary constant.

Calculation:

Consider the function,

  $f^{\text{'}\text{'}}\left(x\right)=\frac{2}{3}{x}^{2/3}$

The antiderivative of $f\left(x\right)$ is,

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

So, by reverse power rule Integrating both sides of equation,

  $\begin{array}{c}{\displaystyle \int F^{`}\left(x\right)}={\displaystyle \int f\left(x\right)}\\ F\left(x\right)={\displaystyle \int f\left(x\right)}\end{array}$

Therefore, integrating the function

  $\begin{array}{c}{f}^{\text{'}\text{'}}\left(x\right)=\frac{2}{3}{x}^{2/3}\\ {\displaystyle \int f^{\text{'}\text{'}}\left(x\right)}dx={\displaystyle \int \frac{2}{3}{x}^{2/3}}dx\\ f^{\text{'}}\left(x\right)=\frac{2}{3}\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}+C\\ f^{\text{'}}\left(x\right)=\frac{2}{5}{x}^{\frac{5}{3}}+C\end{array}$

Again, integrate the function,

  $\begin{array}{c}{\displaystyle \int f^{\text{'}}\left(x\right)dx}={\displaystyle \int \frac{2}{5}{x}^{\frac{5}{3}}+C}dx\\ f\left(x\right)=\frac{2}{5}\frac{x^{\frac{8}{3}}}{\frac{8}{3}}+Cx+D\\ f\left(x\right)=\frac{3}{20}{x}^{\frac{8}{3}}+Cx+D\end{array}$

Thus, the function is $f\left(x\right)=\frac{3}{20}{x}^{\frac{8}{3}}+Cx+D$ .