Chapter 13, Problem 7T

(a)

To determine

The approximate area of the region with five rectangles, lies under the graph of $f\left(x\right)=4-x^2$.

Answer to Problem 7T

The area of each rectangle is 3.56.

Explanation of Solution

Given:

The region lies under the graph of $f\left(x\right)=4-x^2$ and above the interval $0\le x\le 1$ is given below,

Figure (1)

Calculation:

Sketch five rectangles under the curve $f\left(x\right)=4-x^2$ and area of this region is the addition of the areas of each rectangle.

Figure (2)

From the Figure (1), $x_1$ is 0.2, $x_2$ is 0.4, $x_3$ is 0.6, $x_4$ is 0.8, $x_5$ is 1.0.

The formula to find width,

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

Substitute 0 for a, 1 for b and 5 for n in the above formula.

$\begin{array}{c}\Delta x=\frac{1-0}{2}\\ =\frac{1}{2}\\ =0.2\end{array}$

From Figure (2), the area of region is sum of five rectangles

$\begin{array}{c}{R}_5=\left[f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+f\left(x_3\right)\Delta x+f\left(x_4\right)\Delta x+f\left(x_5\right)\Delta x\right]\\ =\Delta x\left[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)+f\left(x_5\right)\right]\end{array}$

Substitute 0.2 for $x_1$, 0.4 for $x_2$, 0.6 for $x_3$, 0.8 for $x_4$ and 1.0 for $x_5$ to find the total area,

$\begin{array}{c}{R}_n=\Delta x\left[f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)+f\left(x_5\right)\right]\\ =\left(0.2\right)\left[\left(4-{\left(0.2\right)}^2\right)+\left(4-{\left(0.4\right)}^2\right)+\left(4-{\left(0.6\right)}^2\right)+\left(4-{\left(0.8\right)}^2\right)+\left(4-{\left(0.2\right)}^2\right)+\left(4-{\left(1\right)}^2\right)\right]\\ =\left(0.2\right)\left[3.96+3.84+3.64+3.36+3\right]\\ =3.56\end{array}$

The area of the region is 3.56.

(b)

To determine

To find: The exact value of the area of the region by using the limit definition of area.

Answer to Problem 7T

The exact value of the area of the region is $\frac{11}{3}$.

Explanation of Solution

Given:

The region lies under the graph of $f\left(x\right)=4-x^2$ and above the interval $0\le x\le 1$ is given below,

Figure (1)

Formula used:

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

$\begin{array}{c}A=\underset{n\to \infty }{\lim }\left[f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+\cdots +f\left(x_n\right)\Delta x\right]\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^{n}f\left(x_k\right)}\Delta x\end{array}$

Where,

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

$x_k=a+k\Delta x$

b and a are the intervals of the region

n is number of rectangles.

Calculation:

The formula to find the width,

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

Substitute 1 for b and 0 for a in the above formula.

$\begin{array}{c}\Delta x=\frac{1-0}{n}\\ =\frac{1}{n}\end{array}$

The formula to calculate right end point is,

$x_k=a+k\Delta x$

Substitute $\frac{1}{n}$ for $\Delta x$ and 0 for a in the above formula.

$\begin{array}{c}{x}_k=0+k\left(\frac{1}{n}\right)\\ =\frac{k}{n}\end{array}$

Substitute $\frac{k}{n}$ for $x_k$ to find the value of $f\left(\frac{k}{n}\right)$,

$\begin{array}{c}f\left(\frac{k}{n}\right)=4-{\left(\frac{k}{n}\right)}^2\\ =\frac{4n^2-k^2}{n^2}\end{array}$

Substitute $\frac{4n^2-k^2}{n^2}$ for $f\left(x_k\right)$, $\frac{1}{n}$ for $\Delta x$ in formula of area $A$,

$\begin{array}{c}A=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\frac{4n^2-k^2}{n^2}}\cdot \frac{1}{n}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\frac{4n^2-k^2}{n^3}}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\frac{4}{n}-\frac{k^2}{n^3}}\\ =\underset{n\to \infty }{\lim }\frac{4}{n}{\displaystyle \sum _{k=1}^{n}1-\underset{n\to \infty }{\lim }\frac{1}{n^3}{\displaystyle \sum _{k=1}^n\frac{k^2}{n^3}}}\end{array}$

Solve above limits,

$\begin{array}{c}A=4-\underset{n\to \infty }{\lim }\frac{1}{n^3}{\displaystyle \sum _{k=1}^n{k}^2}\\ =4-\underset{n\to \infty }{\lim }\frac{1}{n^3}\cdot \frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ =-\underset{n\to \infty }{\lim }\frac{2+\frac{3}{n}+\frac{1}{n^2}}{6}\\ =\frac{11}{3}\end{array}$

Thus the exact value of the area of the region is $\frac{11}{3}$