Chapter 14.5, Problem 20AYU

To determine

To solve: The function $ƒ\left(x\right)=lnx$ is defined on the interval $\left[3,7\right]$,

a. Graph $f$,

indicating the area $A$ under $f$ from 3 to 7.

Answer to Problem 20AYU

a.

Explanation of Solution

Given:

The function $ƒ\left(x\right)=lnx$ is defined on the interval $\left[3,7\right]$.

Calculation:

$ƒ\left(3\right)=1.099;ƒ(\frac{7}{2})=1.253;ƒ\left(4\right)=1.386;ƒ(\frac{9}{2})=1.504;ƒ\left(5\right)=1.609;ƒ(\frac{11}{2})=1.705;ƒ\left(6\right)=1.792;<mtext> </mtext>ƒ(\frac{13}{2})=1.872$

a. Graph $ƒ\left(x\right)=lnx$

To determine

To solve: The function $ƒ\left(x\right)=lnx$ is defined on the interval $\left[3,7\right]$,

b. Approximate the area $A$ by Partition $\left[3,7\right]$ into four subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 20AYU

b. $5.886$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=lnx$ is defined on the interval $\left[3,7\right]$.

Calculation:

$ƒ\left(3\right)=1.099;ƒ(\frac{7}{2})=1.253;ƒ\left(4\right)=1.386;ƒ(\frac{9}{2})=1.504;ƒ\left(5\right)=1.609;ƒ(\frac{11}{2})=1.705;ƒ\left(6\right)=1.792;<mtext> </mtext>ƒ(\frac{13}{2})=1.872$

b. Partition $\left[3,7\right]$ into four subintervals of equal length 1 and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

$A=f\left(3\right)1+f\left(4\right)1+f\left(5\right)1+f\left(6\right)1$

$=1.099\left(1\right)+1.386\left(1\right)+1.609\left(1\right)+1.792\left(1\right)$

$=1.099+1.386+1.609+1.792$

$=5.886$

To determine

To solve: The function $ƒ\left(x\right)=lnx$ is defined on the interval $\left[3,7\right]$,

c. Approximate the area $A$ by Partition $\left[3,7\right]$ into eight subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 20AYU

c. $6.11$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=lnx$ is defined on the interval $\left[3,7\right]$.

Calculation:

$ƒ\left(3\right)=1.099;ƒ(\frac{7}{2})=1.253;ƒ\left(4\right)=1.386;ƒ(\frac{9}{2})=1.504;ƒ\left(5\right)=1.609;ƒ(\frac{11}{2})=1.705;ƒ\left(6\right)=1.792;<mtext> </mtext>ƒ(\frac{13}{2})=1.872$

c. Partition $\left[3,7\right]$ 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

$A=f\left(3\right) \frac{1}{2}+f( \frac{7}{2})\frac{1}{2}+f\left(4\right)( \frac{1}{2})+f( \frac{9}{2})( \frac{1}{2})+f\left(5\right)( \frac{1}{2})+f( \frac{11}{2})( \frac{1}{2})+f\left(6\right)( \frac{1}{2})+f( \frac{13}{2})( \frac{1}{2})$

$=1.099( \frac{1}{2})+\left(1.253\right)( \frac{1}{2})+1.386( \frac{1}{2})+\left(1.504\right)( \frac{1}{2})+\left(1.609\right)( \frac{1}{2})+\left(1.705\right)( \frac{1}{2})+\left(1.792\right)( \frac{1}{2})+\left(1.872\right)( \frac{1}{2})$

$=0.5495+0.6265+0.693+0.752+0.8045+0.8525+0.896+0.936$

$=6.11$

To determine

To solve: The function $ƒ\left(x\right)=lnx$ is defined on the interval $\left[3,7\right]$,

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

Answer to Problem 20AYU

d. $\underset{3}{\overset{7}{{\displaystyle \int }}}ln x dx$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=lnx$ is defined on the interval $\left[3,7\right]$.

Calculation:

$ƒ\left(3\right)=1.099;ƒ(\frac{7}{2})=1.253;ƒ\left(4\right)=1.386;ƒ(\frac{9}{2})=1.504;ƒ\left(5\right)=1.609;ƒ(\frac{11}{2})=1.705;ƒ\left(6\right)=1.792;<mtext> </mtext>ƒ(\frac{13}{2})=1.872$

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

The area $A$ as an integral is $\underset{3}{\overset{7}{{\displaystyle \int }}}ln x dx$.

To determine

To solve: The function $ƒ\left(x\right)=lnx$ is defined on the interval $\left[3,7\right]$,

e. Use a graphing utility to approximate the integral.

Answer to Problem 20AYU

e. $6.326$

Explanation of Solution

Given:

The function $ƒ\left(x\right)=lnx$ is defined on the interval $\left[3,7\right]$.

Calculation:

$ƒ\left(3\right)=1.099;ƒ(\frac{7}{2})=1.253;ƒ\left(4\right)=1.386;ƒ(\frac{9}{2})=1.504;ƒ\left(5\right)=1.609;ƒ(\frac{11}{2})=1.705;ƒ\left(6\right)=1.792;ƒ(\frac{13}{2})=1.872$

e. Use a graphing utility to approximate the integral.

That is evaluate the integral

The value of the integral $\underset{3}{\overset{7}{{\displaystyle \int }}}ln x dx$ is $6.326$.

so the area under the graph of $f$ from 3 to 7 is $6.326$.