Chapter 5.2, Problem 21E

To determine

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

Answer to Problem 21E

The integral of ${\displaystyle \underset{-1}{\overset{5}{\int }}\left(1+3x\right)dx}=42$

Explanation of Solution

Given information:

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

Concept Used:

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

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

Calculation:

The integral is obtained as:

  $\begin{array}{l}{\displaystyle \underset{-1}{\overset{5}{\int }}\left(1+3x\right)dx}\\ \Delta x=\frac{b-a}{n}\\ \Delta x=\frac{5+1}{n}\\ \Delta x=\frac{6}{n}\\ x_i=a+i\Delta x\\ x_i=-1+\frac{6i}{n}\\ f(x_i)=\left(1+3\left(-1+\frac{6i}{n}\right)\right)\end{array}$

And

  $\Delta x=\frac{6}{n}$

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

  $\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(1+3\left(-1+\frac{6i}{n}\right)\right)}\frac{6}{n}$

  $\underset{n\to \infty }{\lim }\left(-2n+\frac{18}{n}\times \frac{n(n+1)}{2}\right)\frac{6}{n}$

  $\underset{n\to \infty }{\lim }\left(-12+54\left(1+\frac{1}{n}\right)\right)=42$