Chapter 14.5, Problem 17AYU

To determine

To solve: The function $ƒ\left(x\right)=\frac{1}{x}$ 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 17AYU

a.

Explanation of Solution

Given:

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

$ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{2}{3};ƒ\left(2\right)=\frac{1}{2};ƒ(\frac{5}{2})=\frac{2}{5};ƒ\left(3\right)=\frac{1}{3};ƒ(\frac{7}{2})=\frac{2}{7};ƒ\left(4\right)=\frac{1}{4};ƒ(\frac{9}{2})=\frac{2}{9}$

Calculation:

a. Graph $ƒ\left(x\right)=\frac{1}{x}$

To determine

To solve: The function $ƒ\left(x\right)=\frac{1}{x}$ 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 17AYU

b. $\frac{25}{12}$

Explanation of Solution

Given:

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

$ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{2}{3};ƒ\left(2\right)=\frac{1}{2};ƒ(\frac{5}{2})=\frac{2}{5};ƒ\left(3\right)=\frac{1}{3};ƒ(\frac{7}{2})=\frac{2}{7};ƒ\left(4\right)=\frac{1}{4};ƒ(\frac{9}{2})=\frac{2}{9}$

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

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

$=1\left(1\right)+\frac{1}{2}\left(1\right)+\frac{1}{3}\left(1\right)+\frac{1}{4} \left(1\right)$

$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{12+6+4+3}{12}$

$=\frac{25}{12}$

To determine

To solve: The function $ƒ\left(x\right)=\frac{1}{x}$ 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 17AYU

c. $\frac{4609}{2520}$

Explanation of Solution

Given:

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

$ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{2}{3};ƒ\left(2\right)=\frac{1}{2};ƒ(\frac{5}{2})=\frac{2}{5};ƒ\left(3\right)=\frac{1}{3};ƒ(\frac{7}{2})=\frac{2}{7};ƒ\left(4\right)=\frac{1}{4};ƒ(\frac{9}{2})=\frac{2}{9}$

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

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

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

$=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}$

$=\frac{1260+840+630+504+420+360+315+280}{2520}=\frac{4609}{2520}$

To determine

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

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

Answer to Problem 17AYU

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

Explanation of Solution

Given:

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

$ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{2}{3};ƒ\left(2\right)=\frac{1}{2};ƒ(\frac{5}{2})=\frac{2}{5};ƒ\left(3\right)=\frac{1}{3};ƒ(\frac{7}{2})=\frac{2}{7};ƒ\left(4\right)=\frac{1}{4};ƒ(\frac{9}{2})=\frac{2}{9}$

Calculation:

d. Express the area $A$ as an integral

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

To determine

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

e. Use a graphing utility to approximate the integral.

Answer to Problem 17AYU

e. $1.609$

Explanation of Solution

Given:

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

$ƒ\left(1\right)=1;ƒ(\frac{3}{2})=\frac{2}{3};ƒ\left(2\right)=\frac{1}{2};ƒ(\frac{5}{2})=\frac{2}{5};ƒ\left(3\right)=\frac{1}{3};ƒ(\frac{7}{2})=\frac{2}{7};ƒ\left(4\right)=\frac{1}{4};ƒ(\frac{9}{2})=\frac{2}{9}$

Calculation:

e. Use a graphing utility to approximate the integral.

That is evaluate the integral.

The value of the integral $\underset{1}{\overset{5}{{\displaystyle \int }}}\frac{1}{x}dx$ is $1.609$,

so the area under the graph of $ƒ$ from 1 to 5 is $1.609$.