Chapter 5.2, Problem 24E

To determine

To Find: The integral of ${\displaystyle \underset{0}{\overset{5}{\int }}\left(1+2x^3\right)dx}$

Answer to Problem 24E

The integral of ${\displaystyle \underset{0}{\overset{5}{\int }}\left(1+2x^3\right)dx}=\frac{635}{2}$

Explanation of Solution

Given information:

  ${\displaystyle \underset{0}{\overset{5}{\int }}\left(1+2x^3\right)dx}$

Concept Used:

If f(x) is integrableon $\left[a,b\right]$ , then

  ${\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx=\underset{n\to \infty }{\lim }}{\displaystyle \sum _{i=1}^{n}f(x_i)\Delta x}$

Where $\Delta x=\frac{b-a}{n}$ and $x_i=a+i\Delta x$

Calculation:

The integral is obtained as:

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{5}{\int }}\left(1+2x^3\right)dx}\\ \Delta x=\frac{b-a}{n}\\ \Delta x=\frac{5-0}{n}\\ \Delta x=\frac{5}{n}\\ x_i=a+i\Delta x\\ x_i=\frac{5i}{n}\\ f(x_i)=1+2{\left(\frac{5i}{n}\right)}^3\end{array}$

And

  $\begin{array}{l}\Delta x=\frac{5}{n}\\ \underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f(x_i)\Delta x}\\ \underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(1+2{\left(\frac{5i}{n}\right)}^3\right)\frac{5}{n}}\\ \underset{n\to \infty }{\lim }\left(5+\frac{1250}{4}\times \frac{n^2{(1+n)}^2}{n^4}\right)\\ \underset{n\to \infty }{\lim }\left(5+\frac{1250}{4}\times \frac{{(1+\frac{1}{n})}^2}{1}\right)\\ \underset{n\to \infty }{\lim }\left(5+\frac{1250}{4}\times \frac{{(1+\frac{1}{n})}^2}{1}\right)=\frac{635}{2}\end{array}$

Hence,

The integral of ${\displaystyle \underset{0}{\overset{5}{\int }}\left(1+2x^3\right)dx}=\frac{635}{2}$