Chapter 11, Problem 16CT

To determine

To calculate: Area bounded by the region of the function.

Answer to Problem 16CT

By the limit process the area bounded by the region is ${\text{12 units}}^2$

Explanation of Solution

Given information: $f\left(x\right)=x+3$ and interval $\left[-2,2\right]$

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

Calculation:

Where, for each positive integer  n , we let  $\Delta x=\frac{b-a}{n}$

Hence, for  $i=1,2,3,\mathrm{........},n,$ we let  $x_i=a+i\Delta x$ b. (These  $x_i$  are the right endpoints of the subintervals.)

Now, ${\displaystyle \underset{-2}{\overset{2}{\int }}\left(x+3\right)dx}$

Find, $\Delta x$

Form each n, we get

  $\Delta x=\frac{b-a}{n}=\frac{2+2}{n}=\frac{4}{n}$

Find, $x_i$

And $x_i=a+i\Delta x=-2+i\frac{4}{n}=-2+\frac{4}{n}i$

Find, $f\left(x_i\right)$

  $\begin{array}{l}f\left(x_i\right)=3+\left(x_i\right)=3+\left(-2+\frac{4}{n}i\right)\\ =1+\frac{4i}{n}\end{array}$

Find and simplify: ${\displaystyle \sum _{i=1}^{n}f\left(x_i\right)\Delta x}$  in order to evaluate the sums.

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)}\Delta x={\displaystyle \sum _{i=1}^n\left(1+\frac{4i}{n}\right)}\times \frac{4}{n}\\ ={\displaystyle \sum _{i=1}^n\frac{4}{n}}+16{\displaystyle \sum _{i=1}^n\frac{i}{n^2}}\\ =\frac{4}{n}{\displaystyle \sum _{i=1}^n\left(1\right)}+\frac{16}{n^2}{\displaystyle \sum _{i=1}^{n}i}\end{array}$

Evaluate the sums:

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)}\Delta x=\frac{4}{n}{\displaystyle \sum _{i=1}^n\left(1\right)}+\frac{16}{n^2}{\displaystyle \sum _{i=1}^{n}i}\\ =\frac{4}{n}\times n+\frac{16}{n^2}\left\{\frac{n\left(n+1\right)}{2}\right\}\\ =4+8\left\{\frac{\left(n+1\right)}{n}\right\}\end{array}$

Now, we need to evaluate the limit as $n\to \infty$

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  $\begin{array}{l}\therefore {\displaystyle \underset{-2}{\overset{2}{\int }}\left(x+3\right)dx}=\underset{n\to \infty }{\lim }\left[4+8\left\{\frac{\left(n+1\right)}{n}\right\}\right]\\ =\underset{n\to \infty }{\lim }4+8\underset{n\to \infty }{\lim }\left(\frac{n}{n}+\frac{1}{n}\right)\\ =\underset{n\to \infty }{\lim }4+8\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\\ =4+8\left(1+0\right)\\ =4+8\\ =12\end{array}$

Area of the bounded region, we have

  ${\displaystyle \underset{-2}{\overset{2}{\int }}\left(x+3\right)dx}=\underset{n\to \infty }{\lim }\left\{4+8\left\{\frac{\left(n+1\right)}{n}\right\}\right\}=12$

Hence, area of the bounded region is ${\text{12 units}}^2$

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