Question

In Exercises 63–68, find a function that has the given derivative and value. In each case you can find the answer with an educated guess-and-check process while thinking about the

chain rule.

$f\text{'}\left(x\right)=x{\left(x^2+1\right)}^4,f\left(0\right)=0$

Short answer

The function is$f\left(x\right)=\frac{{\left(x^2+1\right)}^5}{10}-\frac{1}{10}$

Step-by-step solution

Step 1. Given Information

The given derivative is$f\text{'}\left(x\right)=x{\left(x^2+1\right)}^4,f\left(0\right)=0$

Step 2. Finding antiderivative 

$$\begin{gathered} f\left(x\right)=\int x{\left(x^2+1\right)}^{4}dx \\ f\left(x\right)=\int x{\left(t\right)}^{4}dt \end{gathered}$$

$$\begin{gathered} f\left(x\right)=\frac{1}{2}\int t^{4}dt \\ f\left(x\right)=\frac{1}{2}\frac{t^5}{5}+C \\ f\left(x\right)=\frac{{\left(x^2+1\right)}^5}{10}+C \end{gathered}$$

Step 3. Finding the value of C

Given that $f\left(0\right)=0$

$$\begin{gathered} \frac{{\left(0+1\right)}^5}{10}+C=0 \\ \frac{1}{10}+C=0 \\ C=-\frac{1}{10} \end{gathered}$$

The function become$f\left(x\right)=\frac{{\left(x^2+1\right)}^5}{10}-\frac{1}{10}$