Chapter 11.5, Problem 21E

To determine

To show the approximate area of the region.

Answer to Problem 21E

The required areas are shown below.

Explanation of Solution

Given information:

The function is $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+1$ and the interval is $\left[4,6\right]$ .

  

Approximate area of n rectangles can be calculated as,

  $\mathrm{A}={\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{f}\left(\mathrm{a}+\frac{\left(\mathrm{b}-\mathrm{a}\right)\mathrm{i}}{\mathrm{n}}\right)}\left(\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\right)$

Here,

  $\mathrm{x}=4$ and $\mathrm{y}=6$ .

Hence,

  $\begin{array}{c}\mathrm{A}={\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{f}\left(\mathrm{a}+\frac{\left(\mathrm{b}-\mathrm{a}\right)\mathrm{i}}{\mathrm{n}}\right)}\left(\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\right)\\ ={\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\left(4+\frac{\left(6-4\right)\mathrm{i}}{\mathrm{n}}\right)}\left(\frac{6-4}{\mathrm{n}}\right)\\ ={\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\left(4+\frac{2\mathrm{i}}{\mathrm{n}}+1\right)}\left(\frac{2}{\mathrm{n}}\right)\\ ={\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\left(\frac{34}{\mathrm{n}}+\frac{8{\mathrm{i}}^2}{{\mathrm{n}}^3}+\frac{32\mathrm{i}}{{\mathrm{n}}^2}\right)}\left(\frac{2}{\mathrm{n}}\right)\end{array}$

Using the formula,

  ${\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{i}}^2}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}$ .

  ${\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{i}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}}$ .

Hence,

  $\begin{array}{c}\mathrm{A}=34+\frac{8}{{\mathrm{n}}^3}\cdot \frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{32}{{\mathrm{n}}^2}\cdot \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ =\frac{158{\mathrm{n}}^2+60\mathrm{n}+4}{3{\mathrm{n}}^2}\end{array}$

Substitute 4 for n in the expression $\frac{158{\mathrm{n}}^2+60\mathrm{n}+4}{3{\mathrm{n}}^2}$ .

  $\frac{158\cdot 4^2+60\cdot 4+4}{3\cdot 4^2}=57.57$ .

Substitute 8 for n in the expression $\frac{158{\mathrm{n}}^2+60\mathrm{n}+4}{3{\mathrm{n}}^2}$ .

  $\frac{158\cdot 8^2+60\cdot 8+4}{3\cdot 8^2}=55.1875$ .

Substitute 20 for n in the expression $\frac{158{\mathrm{n}}^2+60\mathrm{n}+4}{3{\mathrm{n}}^2}$ .

  $\frac{158\cdot {20}^2+60\cdot 20+4}{3\cdot {20}^2}=53.67$

Substitute 50 for n in the expression $\frac{158{\mathrm{n}}^2+60\mathrm{n}+4}{3{\mathrm{n}}^2}$ .

  $\frac{158\cdot {50}^2+60\cdot 50+4}{3\cdot {50}^2}=53.0672$

Substitute 100 for n in the expression $\frac{158{\mathrm{n}}^2+60\mathrm{n}+4}{3{\mathrm{n}}^2}$ .

  $\frac{158\cdot {100}^2+60\cdot 100+4}{3\cdot {100}^2}=52.8668$

Substitute $\mathrm{\infty }$ for n in the expression $\frac{158{\mathrm{n}}^2+60\mathrm{n}+4}{3{\mathrm{n}}^2}$ .

  $\underset{\mathrm{n}\to \mathrm{\infty }}{{\displaystyle \lim }}\frac{158{\mathrm{n}}^2+60\mathrm{n}+4}{3{\mathrm{n}}^2}=\frac{458}{3}$

Therefore, the required area are shown above.