Chapter 13.5, Problem 15E

To determine

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

Answer to Problem 15E

The area of the region is 8 square units.

Explanation of Solution

Given information:

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

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 $f\left(x\right)=3x^2$ , $0\le x\le 2$.

Here,

Width:

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

Right end point:

  $\begin{array}{l}{x}_k=a+k\Delta x\\ x_k=0+k\cdot \frac{2}{n}\\ x_k=\frac{2k}{n}\end{array}$

Height:

  $\begin{array}{l}f\left(x_k\right)=3x_k^2\\ f\left(x_k\right)=3{\left(\frac{2k}{n}\right)}^2\\ f\left(x_k\right)=\frac{12k^2}{n^2}\end{array}$

Now, according to definition area is

  $\begin{array}{l}A=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k\right)\Delta x\\ A=\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(\frac{12k^2}{n^2}\cdot \frac{2}{n}\right)\\ A=\underset{n\to \infty }{\lim }\left[\frac{24}{n^3}\sum _{k=1}^n{k}^2\right]\\ A=\underset{n\to \infty }{\lim }\left[\frac{24}{n^3}\cdot \frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]\\ A=\underset{n\to \infty }{\lim }\left[4\cdot \frac{\left(n+1\right)\left(2n+1\right)}{n^2}\right]\\ A=4\underset{n\to \infty }{\lim }\left(\frac{2n^2+3n+1}{n^2}\right)\\ A=4\underset{n\to \infty }{\lim }\left(\frac{\frac{2n^2}{n^2}+\frac{3n}{n^2}+\frac{1}{n^2}}{\frac{n^2}{n^2}}\right)\text{ (Dividing both sides by }{n}^2\text{)}\\ A=4\underset{n\to \infty }{\lim }\left(\frac{2+\frac{3}{n}+\frac{1}{n^2}}{1}\right)\\ A=4\left(\frac{2+\frac{3}{\infty }+\frac{1}{{\infty }^2}}{1}\right)\\ A=4\left(\frac{2+0+0}{1}\right)\text{ (Since }\frac{a}{\infty }=0\text{)}\\ A=4\cdot 2\\ A=8\end{array}$

Hence,

The area of the region is 8 square units.