Chapter 14.5, Problem 18AYU

To determine

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

a. Graph $f$,

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

Answer to Problem 18AYU

a.

Explanation of Solution

Given:

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

Calculation:

$ƒ\left(0\right)=0;ƒ(\frac{1}{2})=0.707;ƒ\left(1\right)=1;ƒ(\frac{3}{2})=1.225;ƒ\left(2\right)=1.414;ƒ(\frac{5}{2})=1.581;ƒ\left(3\right)=1.732;ƒ(\frac{7}{2})=1.871;ƒ\left(4\right)=2$

a. Graph $ƒ\left(x\right)=\sqrt{x}$

To determine

To solve: The function $ƒ\left(x\right)=\sqrt{x}$ 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 18AYU

b. $4.146$

Explanation of Solution

Given:

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

Calculation:

$ƒ\left(0\right)=0;ƒ(\frac{1}{2})=0.707;ƒ\left(1\right)=1;ƒ(\frac{3}{2})=1.225;ƒ\left(2\right)=1.414;ƒ(\frac{5}{2})=1.581;ƒ\left(3\right)=1.732;ƒ(\frac{7}{2})=1.871;ƒ\left(4\right)=2$

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

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

$=0\left(1\right)+1\left(1\right)+1.414\left(1\right)+1.732\left(1\right)$

$=0+1+1.414+1.732$

$=4.146$

To determine

To solve: The function $ƒ\left(x\right)=\sqrt{x}$ 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 18AYU

c. $4.765$

Explanation of Solution

Given:

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

Calculation:

$ƒ\left(0\right)=0;ƒ(\frac{1}{2})=0.707;ƒ\left(1\right)=1;ƒ(\frac{3}{2})=1.225;ƒ\left(2\right)=1.414;ƒ(\frac{5}{2})=1.581;ƒ\left(3\right)=1.732;ƒ(\frac{7}{2})=1.871;ƒ\left(4\right)=2$

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

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

$=0(\frac{1}{2})+\left(0.707\right)(\frac{1}{2})+1(\frac{1}{2})+\left(1.225\right)(\frac{1}{2})+\left(1.414\right)(\frac{1}{2})+\left(1.581\right)(\frac{1}{2})+\left(1.732\right)(\frac{1}{2})\text{ + }\left(1.871\right)(\frac{1}{2})$

$=0+0.3535+0.5+0.6125+0.707+0.7905+0.866+0.9355=4.765$

To determine

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

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

Answer to Problem 18AYU

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

Explanation of Solution

Given:

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

Calculation:

$ƒ\left(0\right)=0;ƒ(\frac{1}{2})=0.707;ƒ\left(1\right)=1;ƒ(\frac{3}{2})=1.225;ƒ\left(2\right)=1.414;ƒ(\frac{5}{2})=1.581;ƒ\left(3\right)=1.732;ƒ(\frac{7}{2})=1.871;ƒ\left(4\right)=2$

d. Express the area $A$ as an integral

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

To determine

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

e. Use a graphing utility to approximate the integral.

Answer to Problem 18AYU

e. $\frac{16}{3}$

Explanation of Solution

Given:

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

Calculation:

$ƒ\left(0\right)=0;ƒ(\frac{1}{2})=0.707;ƒ\left(1\right)=1;ƒ(\frac{3}{2})=1.225;ƒ\left(2\right)=1.414;ƒ(\frac{5}{2})=1.581;ƒ\left(3\right)=1.732;ƒ(\frac{7}{2})=1.871;ƒ\left(4\right)=2$

e. Use a graphing utility to approximate the integral.

That is evaluate the integral

The value of the integral $\underset{0}{\overset{4}{{\displaystyle \int }}}\sqrt{x}dx$ is $\frac{16}{3}$

so the area under the graph of $ƒ$ from 0 to 4 is $\frac{16}{3}$.