Q. 71 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

${\int }_0^{3}x{(x^2+1)}^{1/3}dx$

Step-by-Step Solution

Answer

The solution of the given integral is ${\int }_0^{3}x{(x^2+1)}^{1/3}dx=\frac{3}{8}\left[{10}^{4/3}-1\right]$.

Step 1. Given Information

Solving the given integrals.

${\int }_0^{3}x{(x^2+1)}^{1/3}dx$

Step 2. Using the substitution method.

Let

$$\begin{gathered} u=x^2+1 \\ \frac{du}{dx}=2x \\ du=2xdx \\ \frac{1}{2}du=xdx \end{gathered}$$

Step 3. We will now write the limits of integration ( x = 0   and   x = 3 ) in terms of the new variable u .

When $x=0$, we have

$$\begin{gathered} u=x^2+1 \\ u={\left(0\right)}^2+1 \\ u=0+1 \\ u=1 \end{gathered}$$

When $x=3$, we have

$$\begin{gathered} u=x^2+1 \\ u={\left(3\right)}^2+1 \\ u=9+1 \\ u=10 \end{gathered}$$

Step 4. Using the information in equations, we can change variables completely:

$$\begin{gathered} {\int }_0^{3}x{(x^2+1)}^{1/3}dx=\frac{1}{2}{\int }_1^{10}{u}^{1/3}du \\ {\int }_0^{3}x{(x^2+1)}^{1/3}dx=\frac{1}{2}{\left[\frac{u^{1/3+1}}{1/3+1}\right]}_1^{10} \\ {\int }_0^{3}x{(x^2+1)}^{1/3}dx=\frac{1}{2}{\left[\frac{u^{4/3}}{4/3}\right]}_1^{10} \\ {\int }_0^{3}x{(x^2+1)}^{1/3}dx=\frac{1}{2}·\frac{3}{4}{\left[u^{4/3}\right]}_1^{10} \\ {\int }_0^{3}x{(x^2+1)}^{1/3}dx=\frac{3}{8}\left[{10}^{4/3}-1^{4/3}\right] \\ {\int }_0^{3}x{(x^2+1)}^{1/3}dx=\frac{3}{8}\left[{10}^{4/3}-1\right] \end{gathered}$$