Chapter 4.8, Problem 14E

To determine

To calculate: The most general antiderivative of the function $f\left(x\right)=2\sqrt{x}+6\cos x$ and check answer by differentiation.

Answer to Problem 14E

The most general antiderivative of the function $f\left(x\right)=2\sqrt{x}+6\cos x$ is $F\left(x\right)=\frac{4}{3}{x}^{\frac{3}{2}}+6\sin x+C$ .

Explanation of Solution

Given information:

The function is given as:

  $f\left(x\right)=2\sqrt{x}+6\cos x$

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

Power rule:

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

Calculation:

Consider the function,

  $f\left(x\right)=2\sqrt{x}+6\cos x$

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

  $\begin{array}{c}f\left(x\right)={\displaystyle \int 2\sqrt{x}+6\cos x}dx\\ =2{\displaystyle \int x^{\frac{1}{2}}}dx+6{\displaystyle \int \cos xdx}\\ =2\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+6\sin x+C\end{array}$

Thus, the most general antiderivative of the function is $F\left(x\right)=\frac{4}{3}{x}^{\frac{3}{2}}+6\sin x+C$ .

Now checking the answer by differentiation:

Differentiate the antiderivative function,

  $\begin{array}{c}\frac{d}{dx}F\left(x\right)=\frac{d}{dx}\left(\frac{4}{3}{x}^{\frac{3}{2}}+6\sin x+C\right)\\ F^{`}\left(x\right)=\frac{d}{dx}\left(\frac{4}{3}{x}^{\frac{3}{2}}\right)+\frac{d}{dx}\left(6\sin x\right)+\frac{d}{dx}C\end{array}$

further,

  $f\left(x\right)=2\sqrt{x}+6\cos x$

Thus, answer is correct.