Chapter 4.8, Problem 8E

To determine

To calculate: The most general antiderivative of the function $f\left(x\right)=4\sqrt{x^3}+\sqrt[3]{x^4}$ and check answer by differentiation.

Answer to Problem 8E

The most general antiderivative of the function $f\left(x\right)=4\sqrt{x^3}+\sqrt[3]{x^4}$ is $F\left(x\right)=3x^{\frac{4}{3}}+3\frac{x^{\frac{7}{3}}}{7}+C$ .

Explanation of Solution

Given information:

The function is given as:

  $f\left(x\right)=4\sqrt{x^3}+\sqrt[3]{x^4}$

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

Exponential property:

  $\begin{array}{l}\sqrt[m]{x}=x^{\frac{1}{m}}\\ \text{and}\\ \sqrt[m]{x^n}=x^{\frac{n}{m}}\end{array}$

Power rule:

  $\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n$

Calculation:

Consider the function,

  $f\left(x\right)=4\sqrt{x^3}+\sqrt[3]{x^4}$

Apply exponential rule,

  $f\left(x\right)=4x^{\frac{1}{3}}+x^{\frac{4}{3}}$

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

  $\begin{array}{c}f\left(x\right)=4x^{\frac{1}{3}}+x^{\frac{4}{3}}\\ ={\displaystyle \int 4x^{\frac{1}{3}}}dx+{\displaystyle \int x^{\frac{4}{3}}}dx\\ =4\frac{x^{\frac{4}{3}}}{\frac{4}{3}}+\frac{x^{\frac{7}{3}}}{\frac{7}{3}}+C\\ =3x^{\frac{4}{3}}+3\frac{x^{\frac{7}{3}}}{7}+C\end{array}$

Thus, the most general antiderivative of the function is $F\left(x\right)=3x^{\frac{4}{3}}+3\frac{x^{\frac{7}{3}}}{7}+C$ .

Now checking the answer by differentiation:

Differentiate the antiderivative function,

  $\begin{array}{c}\frac{d}{dx}F\left(x\right)=\frac{d}{dx}\left(3x^{\frac{4}{3}}+3\frac{x^{\frac{7}{3}}}{7}+C\right)\\ F^{`}\left(x\right)=\frac{d}{dx}3x^{\frac{4}{3}}+\frac{d}{dx}3\frac{x^{\frac{7}{3}}}{7}+\frac{d}{dx}C\end{array}$

apply the power rule,

  $f\left(x\right)=4\sqrt{x^3}+\sqrt[3]{x^4}$

So, answer is correct.