Chapter 5.1, Problem 5E

(a)

To determine

To find:

The area under the graph using right endpoints and three rectangles and then area under the graph using right endpoints and six rectangles.

Answer to Problem 5E

The area under the graph using right endpoints and three rectangles is 8 and the area under the graph using right endpoints and six rectangles is 6.875.

Explanation of Solution

Given:

The curve function is $f\left(x\right)=1+x^2$.

The region lies between $x=-1$ and $x=2$. So the limits are $a=-1$ and $b=2$.

Calculation:

Draw the graph for the function $f\left(x\right)=1+x^2$ with three rectangles using the right endpoints as shown in Figure (1).

The expression to find the estimate of the areas of n rectangles $\left(R_n\right)$ using right endpoints is shown below:

$R_n=f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+\mathrm{...}+f\left(x_n\right)\Delta x$ (1)

Here, the right endpoint height of the first rectangle is $f\left(x_1\right)$, the width is $\Delta x$, the right endpoint height of the second rectangle is $f\left(x_2\right)$, and the right endpoint height of n^th rectangle is $f\left(x_n\right)$.

Find the width $\left(\Delta x\right)$ using the relation:

$\Delta x=\frac{b-a}{n}$ (2)

Here, the upper limit is b, the lower limit is a, and the number of rectangles is n.

Find the area estimate for three rectangles with right end points.

Substitute 2 for b, -1 for a, and 3 for n in Equation (2).

$\begin{array}{c}\Delta x=\frac{2+1}{3}\\ =1\end{array}$

From Figure (1), take the right endpoint height of the first rectangle’s $f\left(x_1\right)$ value as 1, the right endpoint height of the second rectangle’s $f\left(x_2\right)$ value as 2, and the right endpoint height of the third rectangle’s $f\left(x_3\right)$ value as 5.

Substitute 3 for n, 1 for $f\left(x_1\right)$, 1 for $\Delta x$, 2 for $f\left(x_2\right)$, and 5 for $f\left(x_3\right)$ in Equation (1).

$\begin{array}{c}{R}_3=\left(1\times 1\right)+\left(2\times 1\right)+\left(5\times 1\right)\\ =1+2+5\\ =8\end{array}$

Therefore, the estimate using right endpoints for $n=3$ is 8.

Draw the graph for the function $f\left(x\right)=1+x^2$ with six rectangles using the right endpoints as shown in Figure (1).

Find the area estimate for six rectangles with right end points.

Substitute 2 for b, -1 for a, and 6 for n in equation (2).

$\begin{array}{c}\Delta x=\frac{2+1}{6}\\ =0.5\end{array}$

From Figure (2), take the right endpoint height of the rectangle’s $f\left(x_1\right)=1.25,f\left(x_2\right)=1,f\left(x_3\right)=1.25,f\left(x_4\right)=2,f\left(x_5\right)=3.25,\text{and }f\left(x_6\right)=5$.

Substitute 6 for n, and the above height of the rectangle’s in equation (1),

$\begin{array}{c}{R}_6=\left(1.25\times 0.5\right)+\left(1\times 0.5\right)+\left(1.25\times 0.5\right)+\left(2\times 0.5\right)+\left(3.25\times 0.5\right)+\left(5\times 0.5\right)\\ =0.625+0.5+0.625+1+1.625+2.5\\ =6.875\end{array}$

Therefore, the estimate using right endpoints for $n=6$ is 6.875.

(b)

To determine

The area under the graph using left endpoints and three rectangles and then area under the graph using left endpoints and six rectangles.

Answer to Problem 5E

The area under the graph using left endpoints and three rectangles is 5 and then area under the graph using left endpoints and six rectangles 5.375.

Explanation of Solution

Draw the graph for the function $f\left(x\right)=1+x^2$ with three rectangles using the left endpoints as shown in Figure (3).

The expression to find the estimate of the areas of n rectangles $\left(L_n\right)$ using left endpoints is shown below:

$L_n=f\left(x_0\right)\Delta x+f\left(x_1\right)\Delta x+\mathrm{...}+f\left(x_{n-1}\right)\Delta x$ (3)

Here, the left endpoint height of the first rectangle is $f\left(x_0\right)$, the width is $\Delta x$, the left endpoint height of the second rectangle is $f\left(x_2\right)$, and the left endpoint height of n^th rectangle is $f\left(x_{n-1}\right)$.

Find the area estimate for three rectangles with left end points:

From Figure (3), take the left endpoint height of the first rectangle’s $f\left(x_0\right)$ value as 2, the left endpoint height of the second rectangle’s $f\left(x_1\right)$ value as 1, and the left endpoint height of the third rectangle’s $f\left(x_2\right)$ value as 2.

Substitute 3 for n, and the above height of the rectangle’s in equation (3),

$\begin{array}{c}{L}_3=\left(2\times 1\right)+\left(1\times 1\right)+\left(2\times 1\right)\\ =2+1+2\\ =5\end{array}$

Therefore, the estimate using left endpoints for $n=3$ is 5

Draw the graph for the function $f\left(x\right)=1+x^2$ with six rectangles using the left endpoints as shown in Figure (4).

Find the area estimate for six rectangles with left end points:

From Figure (4), take the left endpoint height of the rectangle’s $f\left(x_0\right)=2,f\left(x_1\right)=1.25,f\left(x_2\right)=1,f\left(x_3\right)=1.25,f\left(x_4\right)=2,\text{and }f\left(x_5\right)=3.25$.

Substitute 6 for n, 2 for $f\left(x_0\right)$, 0.5 for $\Delta x$, 1.25 for $f\left(x_1\right)$, 1 for $f\left(x_2\right)$, 1.25 for $f\left(x_3\right)$2 for $f\left(x_4\right)$, and 3.25 for $f\left(x_5\right)$ in equation (3).

$\begin{array}{c}{L}_6=\left(2\times 0.5\right)+\left(1.25\times 0.5\right)+\left(1\times 0.5\right)+\left(1.25\times 0.5\right)+\left(2\times 0.5\right)+\left(3.25\times 0.5\right)\\ =1+0.625+0.5+0.625+1+1.625\\ =5.375\end{array}$

Therefore, the estimate using left endpoints for $n=6$ is 5.375.

(c)

To determine

The area under the graph using midpoints and three rectangles and then area under the graph using midpoints and six rectangles.

Answer to Problem 5E

The area under the graph using midpoints is 5.75 and three rectangles and then area under the graph using midpoints and six rectangles is 5.95.

The area under the graph using mid points and three rectangles is 5.75.

The area under the graph using mid points and six rectangles is 5.95.

Explanation of Solution

Draw the graph for the function $f\left(x\right)=1+x^2$ with three rectangles using the midpoints as shown in Figure (5).

The expression to find the estimate of the areas of n rectangles $\left(M_n\right)$ using midpoints is shown below:

$M_n=f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+\mathrm{...}+f\left(x_n\right)\Delta x$ (4)

Here, the midpoint height of the first rectangle is $f\left(x_1\right)$, the width is $\Delta x$, the midpoint height of the second rectangle is $f\left(x_1\right)$, and the midpoint height of n^th rectangle is $f\left(x_n\right)$.

From Figure (5), take the right endpoint height of the rectangle’s $f\left(x_1\right)=1.25,f\left(x_2\right)=1,\text{and}f\left(x_3\right)=3.25$.

Substitute 3 for n, and the above height of the rectangle’s in equation (4),

$\begin{array}{c}{M}_3=\left(1.25\times 1\right)+\left(1.25\times 1\right)+\left(3.25\times 1\right)\\ =1.25+1.5+3.25\\ =5.75\end{array}$

Therefore, the estimate using midpoints for $n=3$ is 5.75.

Draw the graph for the function $f\left(x\right)=1+x^2$ with six rectangles using the midpoints as shown in Figure (6).

Find the area estimate for six rectangles with midpoints:

From Figure (6), take the right endpoint height of the rectangle’s $f\left(x_1\right)=1.6,f\left(x_2\right)=1.1,f\left(x_3\right)=1.1,f\left(x_4\right)=1.6,f\left(x_5\right)=2.5,\text{and }f\left(x_6\right)=4$.

Substitute 6 for n, and the above height of the rectangle’s in equation (4),

$\begin{array}{c}{M}_6=\left(1.6\times 0.5\right)+\left(1.1\times 0.5\right)+\left(1.1\times 0.5\right)+\left(1.6\times 0.5\right)+\left(2.5\times 0.5\right)+\left(4\times 0.5\right)\\ =0.8+0.55+0.55+0.8+1.25+2\\ =5.95\end{array}$

Therefore, the estimate using midpoints for $n=6$ is 5.95.

(d)

To determine

To find:

The best estimate.

Answer to Problem 5E

The midpoint estimate value seems to be the best estimate.

Explanation of Solution

Analyze the best estimate among the estimates:

From part (a), part (b), and part (c), the estimates using right endpoints with three and six rectangles shows an overestimate of the true area and the estimate using left endpoints shows an underestimate of the true area.

The mid estimate of the area using midpoints with three and six rectangles shows the area of each rectangle which appears closer to the true area.

Hence, the mid estimate represents to the best estimate