Chapter 14, Problem 15CT

(a)

To determine

To graph: The function $f$, where $f\left(x\right)=\sqrt{16-x^2}$ is nonnegative and continuous on the interval $\left[0,4\right]$.

Explanation of Solution

Given information:

The function $f\left(x\right)=\sqrt{16-x^2}$.

Graph:

To graph $f\left(x\right)=\sqrt{16-x^2}$ choose random $x$ values and calculate corresponding $y$ values.

Choose $x=0$ and plug in $f\left(x\right)=\sqrt{16-x^2}$ gives $f\left(0\right)=\sqrt{16-{\left(0\right)}^2}=\sqrt{16}=4$

Choose $x=1$ and plug in $f\left(x\right)=\sqrt{16-x^2}$ gives $f\left(1\right)=\sqrt{16-{\left(1\right)}^2}=\sqrt{15}=3.87298$

Choose $x=2$ and plug in $f\left(x\right)=\sqrt{16-x^2}$ gives $f\left(2\right)=\sqrt{16-{\left(2\right)}^2}=\sqrt{12}=3.46410$

Choose $x=4$ and plug in $f\left(x\right)=\sqrt{16-x^2}$ gives $f\left(4\right)=\sqrt{16-{\left(4\right)}^2}=\sqrt{16-16}=0$

So the points on a graph of $f\left(x\right)=\sqrt{16-x^2}$ are $\left(0,4\right),\left(1,3.87298\right),\left(2,3.46410\right)$ and $\left(4,0\right)$.

Now plot these points and draw linepassing through the points

The graph of function $f\left(x\right)=\sqrt{16-x^2}$ is,

Interpretation:

The graph of $f\left(x\right)=\sqrt{16-x^2}$ on interval $\left[0,4\right]$ is one quarter of a circle with radius $4$.

(b)

To determine

Partition of interval $\left[0,4\right]$ into eight subintervals of equal width and choose $u$ as the left endpoint of each subinterval and use the partition to approximate the area under the graph of $f$ from $x=0$ to $x=4$

Answer to Problem 15CT

Solution:

The approximated area under the graph of $f$ from $x=0$ to $x=4$ is $13.359$ sq. units

Explanation of Solution

Given information:

The function, $f\left(x\right)=\sqrt{16-x^2}$.

Explanation:

The interval $\left[0,4\right]$ is partitioned into $8$ subintervals, width of each subinterval is $\frac{4-0}{8}=\frac{1}{2}=0.5$

So, the graph of partition of interval $\left[0,4\right]$ as follows:

Choose $u$ as the left end point

So, $a=u_0=0,u_1=\frac{1}{2},u_2=1,u_3=\frac{3}{2},u_4=2,u_5=\frac{5}{2},u_6=3$ and $u_7=\frac{7}{2}$.

The area, $A$ under the function $f$ from $0$ to $4$ is approximated as follows:

$A=\frac{1}{2}\left[f\left(u_0\right)+f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)+f\left(u_5\right)+f\left(u_6\right)+f\left(u_7\right)\right]$

By substituting values of $u_0,u_1,u_2,u_3,u_4,u_5,u_6,u_7$ and $u_8$

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

$=\frac{1}{2}\left[f\left(0\right)+f\left(0.5\right)+f\left(1\right)+f\left(1.5\right)+f\left(2\right)+f\left(2.5\right)+f\left(3\right)+f\left(3.5\right)\right]$

$=\frac{1}{2}\left[4+\sqrt{16-{\left(0.5\right)}^2}+\sqrt{16-{\left(1\right)}^2}+\sqrt{16-{\left(1.5\right)}^2}+\sqrt{16-{\left(2\right)}^2}+\sqrt{16-{\left(2.5\right)}^2}+\sqrt{16-{\left(3\right)}^2}+\sqrt{16-{\left(3.5\right)}^2}\right]$

$=\frac{1}{2}\left[4+\sqrt{15.75}+\sqrt{15}+\sqrt{13.75}+\sqrt{12}+\sqrt{9.75}+\sqrt{7}+\sqrt{3.75}\right]$

$=\frac{1}{2}\left[4+3.96863+3.87298+3.70810+3.46410+3.12250+2.64575+1.93649\right]$

$=\frac{1}{2}\left[26.71855\right]$

$\approx 13.359$

Hence, approximated area under the graph of $f$ from $x=0$ to $x=4$ is $13.359$ sq. units

(c)

To determine

Exact area of the region under the graph of $f\left(x\right)=\sqrt{16-x^2}$ on the interval $\left[0,4\right]$ and compare it with approximation in part (b).

Answer to Problem 15CT

Solution:

The actual area of the region under the graph of $f\left(x\right)=\sqrt{16-x^2}$ on the interval $\left[0,4\right]$ is $12.56637$ sq. units and the approximated area is slightly greater than actual area.

Explanation of Solution

Given information:

The function, $f\left(x\right)=\sqrt{16-x^2}$.

Explanation:

From part (a) the graph of the function $f\left(x\right)=\sqrt{16-x^2}$ on interval $\left[0,4\right]$ is

The region under the graph of function $f\left(x\right)=\sqrt{16-x^2}$ on interval $\left[0,4\right]$ is one quarter of circle with radius $4$ so the actual area under the graph of $f\left(x\right)=\sqrt{16-x^2}$ on interval $\left[0,4\right]$ is the area of a quarter of circle whose radius is $4$.

Thus, the actual area, $A=\frac{1}{4}\left(\text{area}\text{of}\text{circle}\right)=\frac{1}{4}\left(\pi r^2\right)$.

Substitute $r=4$ in $\frac{1}{4}\left(\pi r^2\right)$,

$=\frac{1}{4}\left(\pi {\left(4\right)}^2\right)$

$=4\pi$

$\approx 12.566$

Thus, the actual area of the region under the graph of $f\left(x\right)=\sqrt{16-x^2}$ on the interval $\left[0,4\right]$ is $12.566$ sq. units

From part (b) the approximated area is $13.359$ sq. units.

By comparing actual area with approximated area it is observed that the approximated area is slightly greater than actual area.