Chapter 4.8, Problem 6E

To determine

To calculate: The most general antiderivative of the function $f\left(x\right)=2x+3x^{1,7}$ and check answer by differentiation.

Answer to Problem 6E

The most general antiderivative of the function $f\left(x\right)=2x+3x^{1,7}$ is $F\left(x\right)==x^2+\frac{x^9}{3}+C$ .

Explanation of Solution

Given information:

The function is given as:

  $f\left(x\right)=2x+3x^{1,7}$

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 rule,

  $x^m{x}^n=x^{m+n}$

Power rule:

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

Calculation:

Consider the function,

  $f\left(x\right)=2x+3x^{1,7}$

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

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

Apply exponential rule,

  $\begin{array}{c}f\left(x\right)=2x+3x^{1+7}\\ =2x+3x^8\end{array}$

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

  $\begin{array}{c}f\left(x\right)=2x+3x^8\\ ={\displaystyle \int 2x}dx+3{\displaystyle \int x^8}dx\\ =2\frac{x^2}{2}+3\frac{x^9}{9}+C\\ =x^2+\frac{x^9}{3}+C\end{array}$

Thus, the most general antiderivative of the function is $F\left(x\right)==x^2+\frac{x^9}{3}+C$ .

Now checking the answer by differentiation:

Differentiate the antiderivative function,

  $\begin{array}{c}\frac{d}{dx}F\left(x\right)=\frac{d}{dx}\left(x^2+\frac{x^9}{3}+C\right)\\ F^{`}\left(x\right)=\frac{d}{dx}{x}^2+\frac{d}{dx}\frac{x^9}{3}+\frac{d}{dx}C\end{array}$

apply the power rule,

  $f\left(x\right)=2x+3x^{1,7}$

So, answer is correct.