Chapter 14.5, Problem 15AYU

To determine

To solve: The function $ƒ\left(x\right)=x^3$ 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 15AYU

a.

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[0,4\right]$.

Calculation:

$ƒ\left(0\right)=0;ƒ(\frac{1}{2})=\frac{1}{8};ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{27}{8};ƒ\left(2\right)=8;ƒ(\frac{5}{2})=\frac{125}{8}$

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

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 $\left[0,4\right]$,

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 15AYU

b. 36

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[0,4\right]$.

Calculation:

$ƒ\left(0\right)=0;ƒ(\frac{1}{2})=\frac{1}{8};ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{27}{8};ƒ\left(2\right)=8;ƒ(\frac{5}{2})=\frac{125}{8}$

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

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

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

$=0\left(1\right)+1\left(1\right)+8\left(1\right)+27\left(1\right)$

$=0+1+8+27$

$=36$

To determine

To solve: The function $ƒ\left(x\right)=x^3$ 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 15AYU

c. 49

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[0,4\right]$.

Calculation:

$ƒ\left(0\right)=0;ƒ(\frac{1}{2})=\frac{1}{8};ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{27}{8};ƒ\left(2\right)=8;ƒ(\frac{5}{2})=\frac{125}{8}$

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

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

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

$=\text{ 0}(\frac{1}{2})+(\frac{1}{8})(\frac{1}{2})+\text{ 1}(\frac{1}{2})+(\frac{27}{8})(\frac{1}{2})+\text{ 8}(\frac{1}{2})+(\frac{125}{8})(\frac{1}{2})+\text{ 27}(\frac{1}{2})\text{ + }\frac{343}{8}(\frac{1}{2})$

$=\text{ 0 }+\frac{1}{16}+\frac{1}{2}+\frac{27}{16}+\text{ 4 }+\frac{125}{16}+\frac{27}{2}+\frac{343}{16}$

$=\frac{\text{0 }+\text{ 1 }+\text{ 8 }+\text{ 27 }+\text{ 64 }+\text{ 125 }+\text{ 216 + 343}}{16}\text{ = }\frac{783}{16}\text{ = }48.9375\text{ = 49}$

To determine

To solve: The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[0,4\right]$,

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

Answer to Problem 15AYU

d. $\underset{0}{\overset{4}{{\displaystyle \int }}}{x}^{3}dx$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[0,4\right]$.

Calculation:

$ƒ\left(0\right)=0;ƒ(\frac{1}{2})=\frac{1}{8};ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{27}{8};ƒ\left(2\right)=8;ƒ(\frac{5}{2})=\frac{125}{8}$

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

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

The area $A$ as an integral is $\underset{0}{\overset{4}{{\displaystyle \int }}}{x}^{3}dx$

To determine

To solve: The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[0,4\right]$,

e. Use a graphing utility to approximate the integral.

Answer to Problem 15AYU

e. 64

Explanation of Solution

Given:

The function $ƒ\left(x\right)=x^3$ is defined on the interval $\left[0,4\right]$.

Calculation:

$ƒ\left(0\right)=0;ƒ(\frac{1}{2})=\frac{1}{8};ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{27}{8};ƒ\left(2\right)=8;ƒ(\frac{5}{2})=\frac{125}{8}$

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

e. Use a graphing utility to approximate the integral.

That is evaluate the integral.

The value of the integral $\underset{0}{\overset{4}{{\displaystyle \int }}}{x}^{3}dx$ is 64, so the area under the graph of $f$ from 0 to 4 is 64.