Chapter 14.5, Problem 14AYU

To determine

To solve: The function $ƒ\left(x\right)=x^2-4$ is defined on the interval $\left[2,6\right]$,

a. Graph $ƒ$,

indicating the area $A$ under $f$ from 2 to 6.

Answer to Problem 14AYU

a.

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^2-4$ is defined on the interval $\left[2,6\right]$.

Calculation:

$ƒ\left(2\right)=2^2-4=0;ƒ(\frac{5}{2})=\frac{25}{4}-4=\frac{9}{4};ƒ\left(3\right)=3^2-4=5;ƒ(\frac{7}{2})=\frac{49}{4}-4=\frac{33}{4};ƒ\left(4\right)=4^2-4=12$;

$ƒ(\frac{9}{2})=\frac{81}{4}–4=\frac{65}{4};ƒ\left(5\right)=25–4=21;ƒ(\frac{11}{2})=\frac{121}{4}–4=\frac{105}{4};ƒ\left(6\right)=36–4=32$

a. Graph $ƒ\left(x\right)=x^2-4$

To determine

To solve: The function $ƒ\left(x\right)=x^2-4$ is defined on the interval $\left[2,6\right]$,

b. Approximate the area $A$ by Partition $\left[2,6\right]$ into four subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 14AYU

b. 38

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^2-4$ is defined on the interval $\left[2,6\right]$.

Calculation:

$ƒ\left(2\right)=2^2-4=0;ƒ(\frac{5}{2})=\frac{25}{4}-4=\frac{9}{4};ƒ\left(3\right)=3^2-4=5;ƒ(\frac{7}{2})=\frac{49}{4}-4=\frac{33}{4};ƒ\left(4\right)=4^2-4=12$;

$ƒ(\frac{9}{2})=\frac{81}{4}–4=\frac{65}{4};ƒ\left(5\right)=25–4=21;ƒ(\frac{11}{2})=\frac{121}{4}–4=\frac{105}{4};ƒ\left(6\right)=36–4=32$

b. Partition $\left[2,6\right]$ into four subintervals of equal length 1 and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

$A=f\left(2\right)1+f\left(3\right)1+f\left(4\right)1+f\left(5\right)1$

$=0\left(1\right)+5\left(1\right)+12\left(1\right)+21\left(1\right)$

$=0+5+12+21$

$=38$

To determine

To solve: The function $ƒ\left(x\right)=x^2-4$ is defined on the interval $\left[2,6\right]$,

c. Approximate the area $A$ by Partition $\left[2,6\right]$ into eight subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 14AYU

c. $\frac{91}{2}$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^2-4$ is defined on the interval $\left[2,6\right]$.

Calculation:

$ƒ\left(2\right)=2^2-4=0;ƒ(\frac{5}{2})=\frac{25}{4}-4=\frac{9}{4};ƒ\left(3\right)=3^2-4=5;ƒ(\frac{7}{2})=\frac{49}{4}-4=\frac{33}{4};ƒ\left(4\right)=4^2-4=12$;

$ƒ(\frac{9}{2})=\frac{81}{4}–4=\frac{65}{4};ƒ\left(5\right)=25–4=21;ƒ(\frac{11}{2})=\frac{121}{4}–4=\frac{105}{4};ƒ\left(6\right)=36–4=32$

c. Partition $\left[2,6\right]$ into eight subintervals of equal length $\frac{1}{2}$ and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

$A=f\left(2\right)(\frac{1}{2})+f(\frac{5}{2})(\frac{1}{2})+f\left(3\right)(\frac{1}{2})+f(\frac{7}{2})(\frac{1}{2})+f\left(4\right)(\frac{1}{2})+f(\frac{9}{2})(\frac{1}{2})+f\left(5\right)\frac{1}{2}+f(\frac{11}{2})(\frac{1}{2})$

$=\text{ 0}(\frac{1}{2})+(\frac{9}{4})(\frac{1}{2})+\text{ 5}(\frac{1}{2})+(\frac{33}{4})(\frac{1}{2})+\text{ 12}(\frac{1}{2})+(\frac{65}{4})(\frac{1}{2})+\text{ 2}1(\frac{1}{2})\text{ + }\frac{105}{4}(\frac{1}{2})$

$=\text{ 0 }+\frac{9}{8}+ \frac{5}{2}+ \frac{33}{8}+\text{ 6 }+\frac{65}{8}+\frac{21}{2}+\frac{105}{8}$

$=\frac{\text{9 }+\text{ 20 }+\text{ 33 }+\text{ 48 }+\text{ 65 }+\text{ 84 }+\text{ 105}}{8}\text{ = }\frac{364}{8}\text{ = }\frac{91}{2}$

To determine

To solve: The function $ƒ\left(x\right)=x^2-4$ is defined on the interval $\left[2,6\right]$,

d. Express the area $A$ as an integral.

Answer to Problem 14AYU

d. $\underset{2}{\overset{6}{{\displaystyle \int }}}\left(x^2- 4\right)dx$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^2-4$ is defined on the interval $\left[2,6\right]$.

Calculation:

$ƒ\left(2\right)=2^2-4=0;ƒ(\frac{5}{2})=\frac{25}{4}-4=\frac{9}{4};ƒ\left(3\right)=3^2-4=5;ƒ(\frac{7}{2})=\frac{49}{4}-4=\frac{33}{4};ƒ\left(4\right)=4^2-4=12$;

$ƒ(\frac{9}{2})=\frac{81}{4}–4=\frac{65}{4};ƒ\left(5\right)=25–4=21;ƒ(\frac{11}{2})=\frac{121}{4}–4=\frac{105}{4};ƒ\left(6\right)=36–4=32$

d. Express the area $A$ as an integral.

The area $A$ as an integral is $\underset{2}{\overset{6}{{\displaystyle \int }}}\left(x^2- 4\right)dx$.

To determine

To solve: The function $ƒ\left(x\right)=x^2-4$ is defined on the interval $\left[2,6\right]$,

e. Use a graphing utility to approximate the integral.

Answer to Problem 14AYU

e. $\frac{160}{3}$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^2-4$ is defined on the interval $\left[2,6\right]$.

Calculation:

$ƒ\left(2\right)=2^2-4=0;ƒ(\frac{5}{2})=\frac{25}{4}-4=\frac{9}{4};ƒ\left(3\right)=3^2-4=5;ƒ(\frac{7}{2})=\frac{49}{4}-4=\frac{33}{4};ƒ\left(4\right)=4^2-4=12$;

$ƒ(\frac{9}{2})=\frac{81}{4}–4=\frac{65}{4};ƒ\left(5\right)=25–4=21;ƒ(\frac{11}{2})=\frac{121}{4}–4=\frac{105}{4};ƒ\left(6\right)=36–4=32$

e. Use a graphing utility to approximate the integral.

That is evaluate the integral

The value of the integral $\underset{2}{\overset{6}{{\displaystyle \int }}}\left(x^2- 4\right)dx$ is $\frac{160}{3}$,

so the area under the graph of $f$ from 2 to 6 is $\frac{160}{3}$.