Chapter 14.5, Problem 13AYU

To determine

To solve: The function $ƒ\left(x\right)=x^2+2$ is defined on the interval $\left[0,4\right]$,

a. Graph $ƒ$.

indicating the area $A$ under $f$ from 0 to 4.

Answer to Problem 13AYU

a.

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^2+2$ is defined on the interval $\left[0,4\right]$.

Calculation:

$ƒ\left(0\right)=0^2+2=2;ƒ(\frac{1}{2})=\frac{1}{4}+2=\frac{9}{4};ƒ\left(1\right)=1+2=3;ƒ(\frac{3}{2})=\frac{9}{4}+2=\frac{17}{4}$

$ƒ\left(2\right)=2^2+2=6;ƒ(\frac{5}{2})=\frac{25}{4}+2=\frac{33}{4};ƒ\left(3\right)=3^2+2=11;ƒ(\frac{7}{2})=\frac{49}{4}+2=\frac{57}{4};ƒ\left(4\right)=4^2+2=18$

a. Graph $ƒ\left(x\right)=x^2+2$

To determine

To solve: The function $ƒ\left(x\right)=x^2+2$ is defined on the interval $\left[0,4\right]$,

b. Approximate the area $A$ by Partition $\left[0,4\right]$ into four subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 13AYU

b. 22

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^2+2$ is defined on the interval $\left[0,4\right]$.

Calculation:

$ƒ\left(0\right)=0^2+2=2;ƒ(\frac{1}{2})=\frac{1}{4}+2=\frac{9}{4};ƒ\left(1\right)=1+2=3;ƒ(\frac{3}{2})=\frac{9}{4}+2=\frac{17}{4}$

$ƒ\left(2\right)=2^2+2=6;ƒ(\frac{5}{2})=\frac{25}{4}+2=\frac{33}{4};ƒ\left(3\right)=3^2+2=11;ƒ(\frac{7}{2})=\frac{49}{4}+2=\frac{57}{4};ƒ\left(4\right)=4^2+2=18$

b. Partition $\left[0,4\right]$ into four subintervals of equal length 1 and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

$A=f\left(0\right)1+f\left(1\right)1+f\left(2\right)1+f\left(3\right)1$

$=2\left(1\right)+3\left(1\right)+6\left(1\right)+11\left(1\right)$

$=2+3+6+11$

$=22$

To determine

To solve: The function $ƒ\left(x\right)=x^2+2$ is defined on the interval $\left[0,4\right]$,

c. Approximate the area $A$ by Partition $\left[0,4\right]$ into eight subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 13AYU

c. $\frac{51}{2}$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^2+2$ is defined on the interval $\left[0,4\right]$.

Calculation:

$ƒ\left(0\right)=0^2+2=2;ƒ(\frac{1}{2})=\frac{1}{4}+2=\frac{9}{4};ƒ\left(1\right)=1+2=3;ƒ(\frac{3}{2})=\frac{9}{4}+2=\frac{17}{4}$

$ƒ\left(2\right)=2^2+2=6;ƒ(\frac{5}{2})=\frac{25}{4}+2=\frac{33}{4};ƒ\left(3\right)=3^2+2=11;ƒ(\frac{7}{2})=\frac{49}{4}+2=\frac{57}{4};ƒ\left(4\right)=4^2+2=18$

c. Partition $\left[0,4\right]$ 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

$A=f\left(0\right)(\frac{1}{2})+f(\frac{1}{2})(\frac{1}{2})+f\left(1\right)(\frac{1}{2})+f(\frac{3}{2})(\frac{1}{2})+f\left(2\right)(\frac{1}{2})+f(\frac{5}{2})(\frac{1}{2})+f\left(3\right)\frac{1}{2}+f(\frac{7}{2})\frac{1}{2}$

$=2(\frac{1}{2})+(\frac{9}{4})(\frac{1}{2})+3(\frac{1}{2})+(\frac{17}{4})(\frac{1}{2})+6(\frac{1}{2})+(\frac{33}{4})(\frac{1}{2})+11(\frac{1}{2})\text{ + }\frac{57}{4}(\frac{1}{2})$

$=1+\frac{9}{8}+ \frac{3}{2}+ \frac{17}{8}+3+\frac{33}{8}+\frac{11}{2}+\frac{57}{8}$

$=\frac{8+9+12+17+24+33+44+57}{8}\text{ = }\frac{204}{8}\text{ = }\frac{51}{2}$

To determine

To solve: The function $ƒ\left(x\right)=x^2+2$ is defined on the interval $\left[0,4\right]$,

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

Answer to Problem 13AYU

d. $\underset{0}{\overset{4}{{\displaystyle \int }}}\left(x^2+2\right)dx$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^2+2$ is defined on the interval $\left[0,4\right]$.

Calculation:

$ƒ\left(0\right)=0^2+2=2;ƒ(\frac{1}{2})=\frac{1}{4}+2=\frac{9}{4};ƒ\left(1\right)=1+2=3;ƒ(\frac{3}{2})=\frac{9}{4}+2=\frac{17}{4}$

$ƒ\left(2\right)=2^2+2=6;ƒ(\frac{5}{2})=\frac{25}{4}+2=\frac{33}{4};ƒ\left(3\right)=3^2+2=11;ƒ(\frac{7}{2})=\frac{49}{4}+2=\frac{57}{4};ƒ\left(4\right)=4^2+2=18$

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

The area $A$ as an integral is $\underset{0}{\overset{4}{{\displaystyle \int }}}\left(x^2+2\right)dx$.

To determine

To solve: The function $ƒ\left(x\right)=x^2+2$ is defined on the interval $\left[0,4\right]$,

e. Use a graphing utility to approximate the integral.

Answer to Problem 13AYU

e. $\frac{88}{3}$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^2+2$ is defined on the interval $\left[0,4\right]$.

Calculation:

$ƒ\left(0\right)=0^2+2=2;ƒ(\frac{1}{2})=\frac{1}{4}+2=\frac{9}{4};ƒ\left(1\right)=1+2=3;ƒ(\frac{3}{2})=\frac{9}{4}+2=\frac{17}{4}$

$ƒ\left(2\right)=2^2+2=6;ƒ(\frac{5}{2})=\frac{25}{4}+2=\frac{33}{4};ƒ\left(3\right)=3^2+2=11;ƒ(\frac{7}{2})=\frac{49}{4}+2=\frac{57}{4};ƒ\left(4\right)=4^2+2=18$

e. Use a graphing utility to approximate the integral.

That is evaluate the integral.

The value of the integral $\underset{0}{\overset{4}{{\displaystyle \int }}}\left(x^2+2\right)dx$ is $\frac{88}{3}$, so the area under the graph of $f$ from 0 to 4 is $\frac{88}{3}$.