Chapter 15.3, Problem 24E

To determine

Find the value of the given integral using the limit.

Answer to Problem 24E

The value of the given integral is $45$ .

Explanation of Solution

Given:

The given integral is ${\displaystyle {\int }_1^4\left(x^2+4x-2\right)dx}$ .

Calculation:

For the given integral $y=x^2+4x-2$ from $x=1$ to $x=4$ .

Apply formula

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)\Delta x}$ , where

  $\Delta x=\frac{b-a}{n}$

  $x_i=a+i\Delta x$

The given integral is

  ${\displaystyle {\int }_1^4\left(x^2+4x-2\right)dx}={\displaystyle {\int }_0^4\left(x^2+4x-2\right)dx}-{\displaystyle {\int }_0^1\left(x^2+4x-2\right)dx}$

For ${\displaystyle {\int }_0^4\left(x^2+4x-2\right)dx}$

For the given region

  $\Delta x=\frac{4-0}{n}=\frac{4}{n}$

  $x_i=0+i\cdot \frac{4}{n}=\frac{4i}{n}$

  $\begin{array}{l}{\displaystyle {\int }_0^4\left(x^2+4x-2\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left[{\left(\frac{4i}{n}\right)}^2+4\left(\frac{4i}{n}\right)-2\right]\frac{4}{n}}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{64i^2}{n^3}+\frac{64i}{n^2}-\frac{8}{n}\right)}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{64i^2}{n^3}\right)}+\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{64i}{n^2}\right)}-\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{8}{n}\right)}\end{array}$

  $\begin{array}{l}=64\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i^2}{n^3}\right)}+64\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i}{n^2}\right)}-8\\ =64\underset{n\to \infty }{\lim }\left(\frac{1^2+2^2+3^2+---+n^2}{n^3}\right)+64\underset{n\to \infty }{\lim }\left(\frac{1+2+3+---+n}{n^2}\right)-8\\ =64\underset{n\to \infty }{\lim }\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6n^3}\right)+64\underset{n\to \infty }{\lim }\left(\frac{n\left(n+1\right)}{2n^2}\right)-8\end{array}$

  $\begin{array}{l}=\frac{64}{6}\underset{n\to \infty }{\lim }\left(\frac{2n^2+3n+1}{n^2}\right)+\frac{64}{2}\underset{n\to \infty }{\lim }\left(\frac{n+1}{n}\right)-8\\ =\frac{32}{3}\underset{n\to \infty }{\lim }\left(\frac{2n^2}{n^2}+\frac{3n}{n^2}+\frac{1}{n^2}\right)+32\underset{n\to \infty }{\lim }\left(\frac{n}{n}+\frac{1}{n}\right)-8\\ =\frac{32}{3}\underset{n\to \infty }{\lim }\left(2+\frac{3}{n}+\frac{1}{n^2}\right)+32\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)-8\end{array}$

  $\begin{array}{l}=\frac{32}{3}\left(\underset{n\to \infty }{\lim }2+\underset{n\to \infty }{\lim }\frac{3}{n}+\underset{n\to \infty }{\lim }\frac{1}{n^2}\right)+32\left(\underset{n\to \infty }{\lim }1+\underset{n\to \infty }{\lim }\frac{1}{n}\right)-8\\ =\frac{32}{3}\left(2+0+0\right)+32\left(1+0\right)-8\end{array}$

  $\begin{array}{l}=\frac{64}{3}+32-8\\ =\frac{64+96-24}{3}\\ =\frac{136}{3}\end{array}$

For ${\displaystyle {\int }_0^1\left(x^2+4x-2\right)dx}$

For the given region

  $\Delta x=\frac{1-0}{n}=\frac{1}{n}$

  $x_i=0+i\cdot \frac{1}{n}=\frac{i}{n}$

  $\begin{array}{l}{\displaystyle {\int }_0^1\left(x^2+4x-2\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left[{\left(\frac{i}{n}\right)}^2+4\left(\frac{i}{n}\right)-2\right]\frac{1}{n}}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i^2}{n^3}+\frac{4i}{n^2}-\frac{2}{n}\right)}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i^2}{n^3}\right)}+\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{4i}{n^2}\right)}-\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{2}{n}\right)}\end{array}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i^2}{n^3}\right)}+4\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(\frac{i}{n^2}\right)}-2\\ =\underset{n\to \infty }{\lim }\left(\frac{1^2+2^2+3^2+---+n^2}{n^3}\right)+4\underset{n\to \infty }{\lim }\left(\frac{1+2+3+---+n}{n^2}\right)-2\\ =\underset{n\to \infty }{\lim }\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6n^3}\right)+4\underset{n\to \infty }{\lim }\left(\frac{n\left(n+1\right)}{2n^2}\right)-2\end{array}$

  $\begin{array}{l}=\frac{1}{6}\underset{n\to \infty }{\lim }\left(\frac{2n^2+3n+1}{n^2}\right)+\frac{4}{2}\underset{n\to \infty }{\lim }\left(\frac{n+1}{n}\right)-2\\ =\frac{1}{6}\underset{n\to \infty }{\lim }\left(\frac{2n^2}{n^2}+\frac{3n}{n^2}+\frac{1}{n^2}\right)+2\underset{n\to \infty }{\lim }\left(\frac{n}{n}+\frac{1}{n}\right)-2\\ =\frac{1}{6}\underset{n\to \infty }{\lim }\left(2+\frac{3}{n}+\frac{1}{n^2}\right)+2\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)-2\end{array}$

  $\begin{array}{l}=\frac{1}{6}\left(\underset{n\to \infty }{\lim }2+\underset{n\to \infty }{\lim }\frac{3}{n}+\underset{n\to \infty }{\lim }\frac{1}{n^2}\right)+2\left(\underset{n\to \infty }{\lim }1+\underset{n\to \infty }{\lim }\frac{1}{n}\right)-2\\ =\frac{1}{6}\left(2+0+0\right)+2\left(1+0\right)-2\end{array}$

  $=\frac{1}{3}$

  ${\displaystyle {\int }_1^4\left(x^2+4x-2\right)dx}=\frac{136}{3}-\frac{1}{3}=\frac{135}{3}=45$

Hence the value of the given integral is $45$ .