Chapter 14.5, Problem 32AYU

To determine

To solve: The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$,

a. Graph $f$.

Answer to Problem 32AYU

a.

Explanation of Solution

Given:

The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$.

Calculation:

$ƒ\left(-1\right)=\sqrt{1-{\left(-1\right)}^2}=0$; $ƒ\left(-0.8\right)=\sqrt{1-{\left(-0.8\right)}^2}=0.36$; $ƒ\left(-0.6\right)=\sqrt{1-{\left(-0.6\right)}^2}=0.64$; $ƒ\left(-0.4\right)=\sqrt{1-{\left(-0.4\right)}^2}=0.84$; $ƒ\left(-0.2\right)=\sqrt{1-{\left(-0.2\right)}^2}=0.96$; $ƒ\left(0\right)=\sqrt{1-{\text{ 0}}^2}=\text{ 1}$; $ƒ\left(0.2\right)=\sqrt{1-\text{ 0}{\text{.2}}^2}=\text{ 0}\text{.96}$; $ƒ\left(0.4\right)=\sqrt{1-\text{ 0}{\text{.4}}^2}=\text{ 0}\text{.84}$; $ƒ\left(0.6\right)=\sqrt{1-\text{ 0}{\text{.6}}^2}=\text{ 0}\text{.64}$; $ƒ\left(0.8\right)=\sqrt{1-\text{ 0}{\text{.8}}^2}=\text{ 0}\text{.36}$; $ƒ\left(1\right)=\sqrt{1-{\text{ 1}}^2}=\text{ 0}$

a. Graph $ƒ\left(x\right)=\sqrt{1-x^2}$

To determine

To solve: The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$,

b. Approximate the area $\text{A}$ under $f$ from $-1$ to 1 into five subintervals of equal length.

Answer to Problem 32AYU

b. 18

Explanation of Solution

Given:

The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$.

Calculation:

$ƒ\left(-1\right)=\sqrt{1-{\left(-1\right)}^2}=0$; $ƒ\left(-0.8\right)=\sqrt{1-{\left(-0.8\right)}^2}=0.36$; $ƒ\left(-0.6\right)=\sqrt{1-{\left(-0.6\right)}^2}=0.64$; $ƒ\left(-0.4\right)=\sqrt{1-{\left(-0.4\right)}^2}=0.84$; $ƒ\left(-0.2\right)=\sqrt{1-{\left(-0.2\right)}^2}=0.96$; $ƒ\left(0\right)=\sqrt{1-{\text{ 0}}^2}=\text{ 1}$; $ƒ\left(0.2\right)=\sqrt{1-\text{ 0}{\text{.2}}^2}=\text{ 0}\text{.96}$; $ƒ\left(0.4\right)=\sqrt{1-\text{ 0}{\text{.4}}^2}=\text{ 0}\text{.84}$; $ƒ\left(0.6\right)=\sqrt{1-\text{ 0}{\text{.6}}^2}=\text{ 0}\text{.64}$; $ƒ\left(0.8\right)=\sqrt{1-\text{ 0}{\text{.8}}^2}=\text{ 0}\text{.36}$; $ƒ\left(1\right)=\sqrt{1-{\text{ 1}}^2}=\text{ 0}$

b. Approximate the area $\text{A}$ under $f$ from $-1$ to 1 into five subintervals of equal length,

The area $\text{A}$ is approximated as,

$\text{A }=f\left(-1\right)0.4+f\left(-0.6\right)1+f\left(-0.2\right)0.4+f\left(0.4\right)0.4+f\left(0.8\right)0.4$

$=0\left(0.4\right)+\left(0.64\right)\left(0.4\right)+\left(0.96\right)\left(0.4\right)+\left(0.84\right)\left(0.4\right)+\left(0.36\right)\left(0.4\right)$

$=0+0.256+0.384+0.336+0.144$

$=1.12$

To determine

To solve: The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$,

c. Approximate the area $\text{A}$ under $f$ from $-1$ to 1 into ten subintervals of equal length.

Answer to Problem 32AYU

c. 12

Explanation of Solution

Given:

The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$.

Calculation:

$ƒ\left(-1\right)=\sqrt{1-{\left(-1\right)}^2}=0$; $ƒ\left(-0.8\right)=\sqrt{1-{\left(-0.8\right)}^2}=0.36$; $ƒ\left(-0.6\right)=\sqrt{1-{\left(-0.6\right)}^2}=0.64$; $ƒ\left(-0.4\right)=\sqrt{1-{\left(-0.4\right)}^2}=0.84$; $ƒ\left(-0.2\right)=\sqrt{1-{\left(-0.2\right)}^2}=0.96$; $ƒ\left(0\right)=\sqrt{1-{\text{ 0}}^2}=\text{ 1}$; $ƒ\left(0.2\right)=\sqrt{1-\text{ 0}{\text{.2}}^2}=\text{ 0}\text{.96}$; $ƒ\left(0.4\right)=\sqrt{1-\text{ 0}{\text{.4}}^2}=\text{ 0}\text{.84}$; $ƒ\left(0.6\right)=\sqrt{1-\text{ 0}{\text{.6}}^2}=\text{ 0}\text{.64}$; $ƒ\left(0.8\right)=\sqrt{1-\text{ 0}{\text{.8}}^2}=\text{ 0}\text{.36}$; $ƒ\left(1\right)=\sqrt{1-{\text{ 1}}^2}=\text{ 0}$

c. Approximate the area $\text{A}$ under $f$ from $-1$ to 1 into ten subintervals of equal length,

The area $\text{A}$ is approximated as,

$\text{A }=f\left(-1\right)0.2+f\left(-0.8\right)0.2+f\left(-0.6\right)0.2+f\left(-0.4\right)0.2+f\left(-0.2\right)0.2+f\left(0\right)0.2+f\left(0.8\right)0.2+f\left(0.6\right)0.2+f\left(0.4\right)0.2+f\left(0.2\right)0.2$

$=\left(0\right)0.2+\left(0.36\right)0.2+\left(0.64\right)0.2+\left(0.84\right)0.2+\left(0.96\right)0.2+1\left(0.2\right)+\left(0.96\right)0.2+\left(0.84\right)0.2+\left(0.64\right)0.2+\left(0.36\right)0.2$

$=1.32$

To determine

To solve: The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$,

d. Express the area as an integral.

Answer to Problem 32AYU

d. $\frac{33}{2}$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$.

Calculation:

$ƒ\left(-1\right)=\sqrt{1-{\left(-1\right)}^2}=0$; $ƒ\left(-0.8\right)=\sqrt{1-{\left(-0.8\right)}^2}=0.36$; $ƒ\left(-0.6\right)=\sqrt{1-{\left(-0.6\right)}^2}=0.64$; $ƒ\left(-0.4\right)=\sqrt{1-{\left(-0.4\right)}^2}=0.84$; $ƒ\left(-0.2\right)=\sqrt{1-{\left(-0.2\right)}^2}=0.96$; $ƒ\left(0\right)=\sqrt{1-{\text{ 0}}^2}=\text{ 1}$; $ƒ\left(0.2\right)=\sqrt{1-\text{ 0}{\text{.2}}^2}=\text{ 0}\text{.96}$; $ƒ\left(0.4\right)=\sqrt{1-\text{ 0}{\text{.4}}^2}=\text{ 0}\text{.84}$; $ƒ\left(0.6\right)=\sqrt{1-\text{ 0}{\text{.6}}^2}=\text{ 0}\text{.64}$; $ƒ\left(0.8\right)=\sqrt{1-\text{ 0}{\text{.8}}^2}=\text{ 0}\text{.36}$; $ƒ\left(1\right)=\sqrt{1-{\text{ 1}}^2}=\text{ 0}$

d. Express the area $\text{A}$ as an integral,

The area $\text{A}$ as an integral is $\underset{-1}{\overset{1}{{\displaystyle \int }}}\sqrt{1-x^2}dx$.

To determine

To solve: The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$,

e. Evaluate the integral using graphing utility.

Answer to Problem 32AYU

e. $\frac{27}{2}$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$.

Calculation:

$ƒ\left(-1\right)=\sqrt{1-{\left(-1\right)}^2}=0$; $ƒ\left(-0.8\right)=\sqrt{1-{\left(-0.8\right)}^2}=0.36$; $ƒ\left(-0.6\right)=\sqrt{1-{\left(-0.6\right)}^2}=0.64$; $ƒ\left(-0.4\right)=\sqrt{1-{\left(-0.4\right)}^2}=0.84$; $ƒ\left(-0.2\right)=\sqrt{1-{\left(-0.2\right)}^2}=0.96$; $ƒ\left(0\right)=\sqrt{1-{\text{ 0}}^2}=\text{ 1}$; $ƒ\left(0.2\right)=\sqrt{1-\text{ 0}{\text{.2}}^2}=\text{ 0}\text{.96}$; $ƒ\left(0.4\right)=\sqrt{1-\text{ 0}{\text{.4}}^2}=\text{ 0}\text{.84}$; $ƒ\left(0.6\right)=\sqrt{1-\text{ 0}{\text{.6}}^2}=\text{ 0}\text{.64}$; $ƒ\left(0.8\right)=\sqrt{1-\text{ 0}{\text{.8}}^2}=\text{ 0}\text{.36}$; $ƒ\left(1\right)=\sqrt{1-{\text{ 1}}^2}=\text{ 0}$

e. Use a graphing utility to approximate the integral,

That is evaluate the integral,

The value of the integral $\underset{-1}{\overset{1}{{\displaystyle \int }}}\sqrt{1-x^2}dx$ is $1.57$, so the area under the graph of $f$ from $-1$ to 1 is $1.57$.

To determine

To solve: The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$,

f. What is the actual area $\text{A}$?

Answer to Problem 32AYU

f. 12

Explanation of Solution

Given:

The function $ƒ\left(x\right)=\sqrt{1-x^2}$ is defined on the interval $\left[-1,1\right]$.

Calculation:

$ƒ\left(-1\right)=\sqrt{1-{\left(-1\right)}^2}=0$; $ƒ\left(-0.8\right)=\sqrt{1-{\left(-0.8\right)}^2}=0.36$; $ƒ\left(-0.6\right)=\sqrt{1-{\left(-0.6\right)}^2}=0.64$; $ƒ\left(-0.4\right)=\sqrt{1-{\left(-0.4\right)}^2}=0.84$; $ƒ\left(-0.2\right)=\sqrt{1-{\left(-0.2\right)}^2}=0.96$; $ƒ\left(0\right)=\sqrt{1-{\text{ 0}}^2}=\text{ 1}$; $ƒ\left(0.2\right)=\sqrt{1-\text{ 0}{\text{.2}}^2}=\text{ 0}\text{.96}$; $ƒ\left(0.4\right)=\sqrt{1-\text{ 0}{\text{.4}}^2}=\text{ 0}\text{.84}$; $ƒ\left(0.6\right)=\sqrt{1-\text{ 0}{\text{.6}}^2}=\text{ 0}\text{.64}$; $ƒ\left(0.8\right)=\sqrt{1-\text{ 0}{\text{.8}}^2}=\text{ 0}\text{.36}$; $ƒ\left(1\right)=\sqrt{1-{\text{ 1}}^2}=\text{ 0}$

f. The actual area under the graph of $ƒ\left(x\right)=\sqrt{1-x^2}$ from $-1$ to 1 is the area of the semi-circle whose radius is 1. The actual area $\text{A}$ is,

Therefore,

$\text{A }=\frac{1}{2} \pi r^2$

$\text{A }=\frac{1}{2}\times \frac{22}{7}\times 1=1.57$