Chapter 14.5, Problem 16AYU

To determine

To solve: The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[1,5\right]$,

a. Graph $f$,

indicating the area $A$ under $f$ from 1 to 5.

Answer to Problem 16AYU

a.

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[1,5\right]$

$ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{27}{8};ƒ\left(2\right)=8;ƒ(\frac{5}{2})=\frac{125}{8};ƒ\left(3\right)=27;ƒ(\frac{7}{2})=\frac{343}{8};ƒ\left(4\right)=64;ƒ(\frac{9}{2})=\frac{729}{8}$

Calculation:

a. Graph $ƒ\left(x\right)=x^3$

To determine

To solve: The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[1,5\right]$,

b. Approximate the area $A$ by Partition $\left[1,5\right]$ into four subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 16AYU

b. 100

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[1,5\right]$

$ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{27}{8};ƒ\left(2\right)=8;ƒ(\frac{5}{2})=\frac{125}{8};ƒ\left(3\right)=27;ƒ(\frac{7}{2})=\frac{343}{8};ƒ\left(4\right)=64;ƒ(\frac{9}{2})=\frac{729}{8}$

Calculation:

b. Partition $\left[1,5\right]$ into four subintervals of equal length 1 and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

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

$=1\left(1\right)+8\left(1\right)+27\left(1\right)+64\left(1\right)$

$=1+8+27+64$

$=100$

To determine

To solve: The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[1,5\right]$,

c. Approximate the area $A$ by Partition $\left[1,5\right]$ into eight subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 16AYU

c. $\frac{253}{2}$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[1,5\right]$

$ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{27}{8};ƒ\left(2\right)=8;ƒ(\frac{5}{2})=\frac{125}{8};ƒ\left(3\right)=27;ƒ(\frac{7}{2})=\frac{343}{8};ƒ\left(4\right)=64;ƒ(\frac{9}{2})=\frac{729}{8}$

Calculation:

c. Partition $\left[1,5\right]$ 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

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

$= 1(\frac{1}{2})+(\frac{27}{8})(\frac{1}{2})+8(\frac{1}{2})+(\frac{125}{8})(\frac{1}{2})+27( \frac{1}{2})+\frac{343}{8}( \frac{1}{2})+64(\frac{1}{2})+(\frac{729}{8})(\frac{1}{2})$

$=\frac{1}{2}+\frac{27}{6}+4+\frac{125}{16}+\frac{27}{2}+\frac{343}{16}+32+\frac{729}{16}$

$= \frac{8+27+64+125+216+343+512+729}{16}=\frac{2024}{16}=\frac{253}{2}$

To determine

To solve: The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[1,5\right]$,

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

Answer to Problem 16AYU

d. $\underset{1}{\overset{5}{{\displaystyle \int }}}{x}^{3}dx$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[1,5\right]$

$ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{27}{8};ƒ\left(2\right)=8;ƒ(\frac{5}{2})=\frac{125}{8};ƒ\left(3\right)=27;ƒ(\frac{7}{2})=\frac{343}{8};ƒ\left(4\right)=64;ƒ(\frac{9}{2})=\frac{729}{8}$

Calculation:

d. Express the area $A$ as an integral

The area $A$ as an integral is $\underset{1}{\overset{5}{{\displaystyle \int }}}{x}^{3}dx$

To determine

To solve: The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[1,5\right]$,

e. Use a graphing utility to approximate the integral.

Answer to Problem 16AYU

e. 156

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[1,5\right]$

$ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{27}{8};ƒ\left(2\right)=8;ƒ(\frac{5}{2})=\frac{125}{8};ƒ\left(3\right)=27;ƒ(\frac{7}{2})=\frac{343}{8};ƒ\left(4\right)=64;ƒ(\frac{9}{2})=\frac{729}{8}$

Calculation:

e. Use a graphing utility to approximate the integral.

That is evaluate the integral

The value of the integral $\underset{1}{\overset{5}{{\displaystyle \int }}}{x}^{3}dx$ is 156,

so the area under the graph of $ƒ$ from 1 to 5 is 156.