Chapter 12.5, Problem 37E

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 37E

The area of region is $\text{3}\text{square}\text{units}$

Explanation of Solution

Given information:

The given function is,

  $f\left(x\right)=4x+1$

Given intervals $\left[0,1\right]$

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,

  $f\left(x\right)=4x+1$

The area of the region bounded by the graph of $f\left(x\right)=4x+1$ and the x-axis between $x=0$ and $x=1$

The dimensions of the rectangles;

$\begin{array}{c}width;\frac{b-a}{n}=\frac{1-0}{n}\\ =\frac{1}{n}\end{array}$

And,

$\begin{array}{c}\text{Height}\text{;}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(0+\frac{\left(1-0\right)i}{n}\right)\\ =f\left(\frac{i}{n}\right)\end{array}$

Given function; $f\left(x\right)=4x+1$

  $\begin{array}{c}f\left(x\right)=4\left(\frac{1i}{n}\right)+1\\ =\frac{4i}{n}+1\\ =\frac{4i+n}{n}\end{array}$

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

  $\begin{array}{c}\text{A=}\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{\text{i=1}}^{\text{n}}\text{f}\left(\text{a+}\frac{\left(\text{b-a}\right)\text{i}}{\text{n}}\right)\left(\frac{\text{b-a}}{\text{n}}\right)}\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\left(\frac{4i+n}{n}\right)\left(\frac{1}{n}\right)}\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\frac{4i}{n^2}+\frac{1}{n}}\right)\end{array}$

Apply Summation formula and properties.

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

Now,

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left(\frac{4}{n^2}{\displaystyle \sum _{i=1}^n\left(i\right)}+\frac{1}{n}{\displaystyle \sum _{i=1}^{n}1}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{4}{n^2}\times \frac{n\left(n+1\right)}{2}+1\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{2\left(n+1\right)}{n}+1\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{3n+2}{n}\right)\end{array}$

Finally the exact area by taking the limit as $n$ approaches $\infty$

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left[\frac{3n+2}{n}\right]\\ =3\end{array}$

Therefore, the exact area is $\text{3}\text{square}\text{units}$