Chapter 12, Problem 16CT

To determine

Find the area of the region bounded by the graph of the function.

Answer to Problem 16CT

The area of region is $\boxed{A=8}$ square unit.

Explanation of Solution

Given information:

Use the limit process to find the area of the region bounded by the graph of the function and the -axis over the specified interval.

  $f\left(x\right)=x+2$

  $\begin{array}{l}\mathrm{int}erval\\ \left[-2,2\right]\end{array}$

Calculation:

Consider the given function and corresponding interval ,

  $f\left(x\right)=x+2$

  $\begin{array}{l}\mathrm{int}erval\\ \left[-2,2\right]\end{array}$

From the interval,

  $\begin{array}{l}a=-2\\ b=2\end{array}$

Now calculate width and height by using standard formula,

  $\begin{array}{l}Width=\frac{b-a}{n}\\ Height=f\left(a+\frac{\left(b-a\right)i}{n}\right)\end{array}$

Substitute the given values ,

  $Width=\frac{2-\left(-2\right)}{n}=\frac{4}{n}$

  $Height=f\left(-2+\frac{4i}{n}\right)=\frac{4i}{n}$

Now find the area with the help of formula.

  $\begin{array}{l}A={\displaystyle \sum _{i=1}^{n}f\left(a+\frac{\left(b-a\right)i}{n}\right)}\left(\frac{b-a}{n}\right)\\ ={\displaystyle \sum _{i=1}^n\left(\frac{4i}{n}\right)}\left(\frac{4}{n}\right)\\ =\frac{16}{n^2}{\displaystyle \sum _{i=1}^{n}i}\end{array}$

Apply standard formula of summation,

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{n}\\ hence,\\ A=\frac{16}{n^2}\left(\frac{n\left(n+1\right)}{2}\right)\\ =8\left(\frac{n^2+n}{n^2}\right)\end{array}$

Now divide numerator and denominator by $n^2$ ,

  $\begin{array}{l}A=\underset{n\to \infty }{\lim }8\left(\frac{1+\frac{1}{n}}{1}\right)\\ n\to \infty \\ \boxed{A=8}\end{array}$

Hence the area of region is $\boxed{A=8}$ square unit.