Chapter 11, Problem 109CR

To determine

To evaluate: the area of region bounded by the graph of the function.

Answer to Problem 109CR

The area of region bounded by the graph of the function is 21.

Explanation of Solution

Given:

  $f\left(x\right)=x^2+4,\left[0,3\right]$

It is given that

  $f\left(x\right)=x^2+4,\left[0,3\right]$

So,

  $\begin{array}{c}\Delta x=\frac{b-a}{n}\\ =\frac{3-0}{n}\\ =\frac{3}{n}\end{array}$

Now,

  $\begin{array}{c}{x}_i=0+i\frac{3}{n}\\ x_i=\frac{3}{n}i\\ f\left(x_i\right)=x^2+4\\ ={\left(\frac{3}{n}i\right)}^2+4\\ =\frac{9}{n^2}{i}^2+4\end{array}$

So,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)}\Delta x={\displaystyle \sum _{i=1}^n\left(\frac{9}{n^2}{i}^2+4\right)}\left(\frac{3}{n}\right)\\ ={\displaystyle \sum _{i=1}^n\left(\frac{27}{n^3}\right)i^2}+{\displaystyle \sum _{i=1}^n\left(\frac{12}{n}\right)}\\ =\frac{27}{n^3}\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)+\frac{12}{n}\left(n\right)\\ =\frac{27}{1}\left(\frac{1\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\right)+\frac{12}{1}\end{array}$

Now, finding the limit,

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{27}{1}\left(\frac{1\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\right)+\frac{12}{1}=27\left(\frac{2}{6}\right)+12\\ =9+12\\ =21\end{array}$

So, the area of region bounded by the graph of the function is 21.