Chapter 5.1, Problem 4E

(a)

To determine

To Analyze:

Whether the estimate is an underestimate or an overestimate using right endpoints.

Answer to Problem 4E

The estimate is an overestimate for the function $f\left(x\right)=\sin x$ with four approximating rectangles and right endpoints.

Explanation of Solution

Given information:

The curve function is $f\left(x\right)=\sin x$.

The region lies between $x=0$ and $x=\frac{\pi }{2}$. So, the limits are $a=0$ and $b=\frac{\pi }{2}$.

The number of rectangles $n=4$.

Draw the graph for the function $f\left(x\right)=\sin x$ as shown below in Figure (1):

Draw four approximating rectangles using right end points for the function $f\left(x\right)=\sin x$ as shown below in Figure (2):

The expression to find the lower estimate of areas of n rectangles $\left(R_n\right)$ 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.

Substitute $\frac{\pi }{2}$ for b, 0 for a, and 4 for n in Equation (2).

$\begin{array}{c}\Delta x=\frac{\frac{\pi }{2}-0}{4}\\ =\frac{\pi }{8}\end{array}$

Refer Figure (2).

Take the right endpoint height of the first rectangle $f\left(x_1\right)$ value as 0.38, the right endpoint height of the second rectangle $f\left(x_2\right)$ value as 0.7, the right endpoint height of the third rectangle $f\left(x_3\right)$ value as 0.93, and the right endpoint height of the fourth rectangle $f\left(x_4\right)$ value as 1.

Substitute 4 for n, 0.38for $f\left(x_1\right)$, $\frac{\pi }{8}$ for $\Delta x$, 0.7 for $f\left(x_2\right)$, 0.93 for $f\left(x_3\right)$, and 1 for $f\left(x_4\right)$ in Equation (1).

$\begin{array}{c}{R}_4=\left(0.38\times \frac{\pi }{8}\right)+\left(0.7\times \frac{\pi }{8}\right)+\left(0.93\times \frac{\pi }{8}\right)+\left(1\times \frac{\pi }{8}\right)\\ =0.149+0.275+0.365+0.393\\ =\mathrm{1.182.}\end{array}$

Refer Figure (2).

The function curve as $f\left(x\right)=\sin x$ is the increasing curve in the interval between $x=0$ and $x=\frac{\pi }{2}$. The intervals are represented as right end points.

Hence, $R_4$ is an over estimate.

(b)

To determine

To Analyze:

Whether the estimate is an underestimate or an overestimate using right endpoints.

Answer to Problem 4E

The estimate is an underestimate for the function $f\left(x\right)=\sin x$ with four approximating rectangles and left endpoints.

Explanation of Solution

Draw four approximating rectangles using left end points for the function $f\left(x\right)=\sin x$ as shown below in Figure (3):

The expression to find the upper estimate of areas of 4 rectangles $\left(L_4\right)$ is shown below:

$L_4=R_4+f\left(x_0\right)\Delta x-f\left(x_3\right)\Delta x$ (3)

Here, the lower estimate using right endpoints for $n=4$ is $R_4$, left endpoint height of left uppermost rectangle is $f\left(x_0\right)$, and the left endpoint height of the lowermost right rectangle is $f\left(x_3\right)$.

Refer Figure (3).

Take the left endpoint height of the uppermost left rectangle $f\left(x_0\right)$ value as 0.99 and the left endpoint height $f\left(x_3\right)$ value as 0.58.

Substitute 1.182 for $R_4$, 0.38 for $f\left(x_0\right)$, $\frac{\pi }{8}$ for $\Delta x$, and 0.94 for $f\left(x_3\right)$ in Equation (3).

$\begin{array}{c}{L}_4=R_4+f\left(x_0\right)\Delta x-f\left(x_3\right)\Delta x\\ =1.182+\left(0.38\times \frac{\pi }{8}\right)-\left(0.94\times \frac{\pi }{8}\right)\\ =1.182+0.149-0.369\\ =\mathrm{0.962.}\end{array}$

Refer Figure (3).

The function curve as $f\left(x\right)=\sin x$ is increasing in the interval between $x=0$ and $x=\frac{\pi }{2}$ and the intervals are represented as left end points.

Hence, $L_4$ is an underestimate.