Chapter 13.5, Problem 13E

To determine

To find : The area of the region under the curve and check your answer by sketching the region and using geometry.

Answer to Problem 13E

The area of the region is $\frac{75}{2}$ square units.

Explanation of Solution

Given information:

The given function is $y=3x$ , $0\le x\le 5$.

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:

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

Let $f\left(x\right)=3x$ , $0\le x\le 5$.

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:

  $\begin{array}{l}f\left(x_k\right)=3x_k\\ f\left(x_k\right)=3\cdot \frac{5k}{n}\\ f\left(x_k\right)=\frac{15k}{n}\end{array}$

Now, according to definition area is

  $\begin{array}{l}A=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k\right)\Delta x\\ A=\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{15k}{n}\cdot \frac{5}{n}\\ A=\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{75k}{n^2}\\ A=\underset{n\to \infty }{\lim }\frac{75}{n^2}\sum _{k=1}^{n}k\\ A=\underset{n\to \infty }{\lim }\left(\frac{75}{n^2}\cdot \frac{n\left(n+1\right)}{2}\right)\\ A=\underset{n\to \infty }{\lim }\left(\frac{75}{2}\cdot \frac{\left(n+1\right)}{n}\right)\\ A=\frac{75}{2}\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\\ A=\frac{75}{2}\left(1+\frac{1}{\infty }\right)\\ A=\frac{75}{2}\left(1+0\right)\text{ (Since }\frac{a}{\infty }=0\text{)}\\ A=\frac{75}{2}\end{array}$

Now draw the graph of $y=3x$ , $0\le x\le 5$

  

The shaded region is the required region. Since it a triangle, therefore area is

  $\begin{array}{l}A=\frac{1}{2}\cdot 5\cdot 15\\ A=\frac{75}{2}\end{array}$

Hence,

The area of the region is $\frac{75}{2}$ square units.