Chapter 7.2, Problem 59E

To determine

To determine the integral ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{t^5+2t}(5t^4+2)dt}$ by substitution.

Answer to Problem 59E

The value of the integral ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{t^5+2t}(5t^4+2)dt}$ is $\frac{2}{9}\left[3^{3/2}\right]$

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{t^5+2t}(5t^4+2)dt}$

Formula used: ${\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C}$

Calculation : Put $t^5+2t=u$ so that $\left(5t^4+2\right)dt=du$

Also, when $t=0;u=0$ and when $t=1;u=3$

The given integral becomes as:

  ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{t^5+2t}(5t^4+2)dt}={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{u}du}$

  $=\frac{1}{3}{\left[\frac{u^{3/2}}{3/2}\right]}_0^3=\frac{2}{9}\left[3^{3/2}-0\right]=$

Hence, ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{t^5+2t}(5t^4+2)dt}=\frac{2}{9}\left[3^{3/2}\right]$