Chapter 13.5, Problem 17E

To determine

To find : The area of the region under the graph of f over the given interval.

Answer to Problem 17E

The area of the region is $\frac{665}{4}$ square units.

Explanation of Solution

Given information:

The given function is $f\left(x\right)=x^3+2$ , $0\le x\le 5$.

Calculation :

The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles:

  $\begin{array}{l}A=\underset{n\to \infty }{\lim }{R}_n=\underset{n\to \infty }{\lim }\left[f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+\mathrm{......}+f\left(x_n\right)\Delta x\right]\text{ or }\\ A=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k\right)\Delta x\end{array}$

The function is $f\left(x\right)=x^3+2$ , $0\le x\le 5$.

Here,

Width:

  $\Delta x=\frac{b-a}{n}=\frac{5-0}{n}=\frac{5}{n}$

Right end point:

  $\begin{array}{l}{x}_k=a+k\Delta x\\ x_k=0+k\cdot \frac{5}{n}\\ x_k=\frac{5k}{n}\end{array}$

Height:

  $\begin{array}{l}f\left(x_k\right)=x_k^3+2\\ f\left(x_k\right)={\left(\frac{5k}{n}\right)}^3+2\\ f\left(x_k\right)=\frac{125k^3}{n^3}+2\end{array}$

Now, according to definition area is

  $\begin{array}{l}A=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k\right)\Delta x\\ A=\underset{n\to \infty }{\lim }\sum _{k=1}^n\left[\left(\frac{125k^3}{n^3}+2\right)\cdot \frac{5}{n}\right]\\ A=\underset{n\to \infty }{\lim }\left(\sum _{k=1}^n\frac{625k^3}{n^4}+\sum _{k=1}^n\frac{10}{n}\right)\\ A=\underset{n\to \infty }{\lim }\left(\frac{625}{n^4}\sum _{k=1}^n{k}^3+10\right)\\ A=10+\underset{n\to \infty }{\lim }\left[\frac{625}{n^4}\cdot \frac{n^2{\left(n+1\right)}^2}{4}\right]\\ A=10+\underset{n\to \infty }{\lim }\left[\frac{625}{n^4}\cdot \frac{n^2\left(n^2+2n+1\right)}{4}\right]\\ A=10+\underset{n\to \infty }{\lim }\left(\frac{625}{n^4}\cdot \frac{n^4+2n^3+n^2}{4}\right)\\ A=10+\frac{625}{4}\underset{n\to \infty }{\lim }\left(1+\frac{2}{n}+\frac{1}{n^2}\right)\\ A=10+\frac{625}{4}\left(1+\frac{2}{\infty }+\frac{1}{{\infty }^2}\right)\\ A=10+\frac{625}{4}\left(1+0+0\right)\text{ (Since }\frac{a}{\infty }=0\text{)}\\ A=\frac{40+625}{4}\\ A=\frac{665}{4}\end{array}$

Hence,

The area of the region is $\frac{665}{4}$ square units.