Chapter 11.5, Problem 34E

To determine

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

Answer to Problem 34E

By the limit process the area bounded by the region is ${\text{6 units}}^2$

Explanation of Solution

Given information: $f\left(x\right)=x+2$ and interval $\left[0,2\right]$

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, ${\displaystyle \underset{0}{\overset{2}{\int }}\left(x+2\right)dx}$

Find, $\Delta x$

Form each n, we get

  $\Delta x=\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n}$

Find, $x_i$

And $x_i=a+i\Delta x=0+i\frac{2}{n}=i\frac{2}{n}$

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

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

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

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

Evaluate the sums:

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

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

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  $\begin{array}{l}\therefore {\displaystyle \underset{0}{\overset{2}{\int }}\left(x+2\right)dx}=\underset{n\to \infty }{\lim }\left[4+2\left\{\frac{\left(n+1\right)}{n}\right\}\right]\\ =\underset{n\to \infty }{\lim }4+2\underset{n\to \infty }{\lim }\left(\frac{n}{n}+\frac{1}{n}\right)\\ =\underset{n\to \infty }{\lim }4+2\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\\ =4+2\left(1+0\right)\\ =4+2\\ =6\end{array}$

Area of the bounded region, we have

  ${\displaystyle \underset{0}{\overset{2}{\int }}\left(x+2\right)dx}=\underset{n\to \infty }{\lim }\left\{4+2\left\{\frac{\left(n+1\right)}{n}\right\}\right\}=6$

Hence, area of the bounded region is ${\text{6 units}}^2$

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