Chapter 12.5, Problem 40E

To determine

To find: The area of region bounded by the graph of $f$ and the x-axis over the specified interval using the limit process.

Answer to Problem 40E

The area of region is $25.5\text{square}\text{units}$

Explanation of Solution

Given information:

The given function is,

  $f\left(x\right)=3x-4$

Given intervals $\left[2,5\right]$

Concept used:

The area of a region;

Let $f$ be continuous and nonnegative on the interval $\left[a,b\right]$ The area A of the region bounded by the graph of $f$ , the $\text{x-axis}$ ,and the vertical lines $x=a$ and $x=b$ is given by

  $A=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\underset{\text{height}}{\underbrace{\left(a+\frac{\left(b-a\right)i}{n}\right)}}}\underset{\text{width}}{\underbrace{\left(\frac{b-a}{n}\right)}}$

The given function is,

  $f\left(x\right)=3x-4$

The area of the region bounded by the graph of $f\left(x\right)=3x-4$ and the x-axis between $x=2$ and $x=5$

The dimensions of the rectangles;

$\begin{array}{c}width;\frac{b-a}{n}=\frac{5-2}{n}\\ =\frac{3}{n}\end{array}$

And,

  $\begin{array}{c}\text{Height}\text{;}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(2+\frac{\left(5-2\right)i}{n}\right)\\ =f\left(2+\frac{3i}{n}\right)\end{array}$

Given function; $f\left(x\right)=3x-4$

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

Approximate the areas as the sum of the areas of $n$ rectangles.

  $\begin{array}{c}\text{A=}\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{\text{i=1}}^{\text{n}}\text{f}\left(\text{a+}\frac{\left(\text{b-a}\right)\text{i}}{\text{n}}\right)\left(\frac{\text{b-a}}{\text{n}}\right)}\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\left(\frac{9i+2n}{n}\right)\left(\frac{3}{n}\right)}\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\frac{27i}{n^2}+\frac{6}{n}}\right)\end{array}$

Apply Summation formula and properties.

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

Now,

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left(\frac{27}{n^2}{\displaystyle \sum _{i=1}^n\left(i\right)}+\frac{1}{n}{\displaystyle \sum _{i=1}^{n}6}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{27}{n^2}\times \frac{n\left(n+1\right)}{2}+6\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{27\left(n+1\right)}{2n}+6\right)\end{array}$

Finally the exact area by taking the limit as

  $n$ approaches $\infty$

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left[\frac{27\left(n+1\right)}{2n}+6\right]\\ =19.5+6\\ =25.5\end{array}$

Therefore, the exact area is $25.5\text{square}\text{units}$