Chapter 5.1, Problem 23E

(a)

To determine

To find: The relation between area, lower sum, and upper sum for an increasing function f.

Answer to Problem 23E

The relation between area, lower sum, and upper sum for a function is $\underset{\_}{L_n<A<R_n}$.

Explanation of Solution

Given information:

The curve f is an increasing continuous function.

Describe the relation between $L_n$, A, and $R_n$ as below:

• The increasing function f represents the lower sum $\left(L_n\right)$ and an upper sum $\left(R_n\right)$. So $L_n$ is an underestimate of area (A) and $R_n$ is an overestimate of area (A).
• The value of $R_n$ is more than the area A and the value of $L_n$ is less than the area A.

Therefore, the terms are related as $\underset{\_}{L_n<A<R_n}$.

(b)

To determine

To prove: The relation $R_n-L_n=\frac{b-a}{n}\left[f\left(b\right)-f\left(a\right)\right]$.

Answer to Problem 23E

The relation $R_n-L_n=\frac{b-a}{n}\left[f\left(b\right)-f\left(a\right)\right]$ is proved.

Explanation of Solution

The expression for upper estimate of area 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 of the interval 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 $f\left(x_n\right)$.

The expression for lower estimate of area is shown below:

$L_n=f\left(x_0\right)\Delta x+f\left(x_1\right)\Delta x+\mathrm{...}+f\left(x_{n-1}\right)\Delta x$ (2)

Here, the left endpoint height of the first rectangle is $f\left(x_0\right)$, the left endpoint height of the second rectangle is $f\left(x_1\right)$, and the left endpoint height of n^th rectangle is $f\left(x_{n-1}\right)$.

Subtract Equation (2) from Equation (1) as shown below:

$\begin{array}{c}{R}_n-L_n=\left[\begin{array}{l}\left[f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+\mathrm{...}+f\left(x_n\right)\Delta x\right]\\ -\left[f\left(x_0\right)\Delta x+f\left(x_1\right)\Delta x+\mathrm{...}+f\left(x_{n-1}\right)\Delta x\right]\end{array}\right]\\ =f\left(x_n\right)\Delta x-f\left(x_0\right)\Delta x\\ =\Delta x\left[f\left(x_n\right)-f\left(x_0\right)\right]\end{array}$ (3)

The width of the interval $\left(\Delta x\right)$ is expressed as shown below:

$\Delta x=\frac{b-a}{n}$ (4)

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

Substitute $b$ for $x_n$, a for $x_0$, and $\frac{b-a}{n}$ for $\Delta x$ in Equation (3).

$R_n-L_n=\frac{b-a}{n}\left[f\left(b\right)-f\left(a\right)\right]$

Hence, the relation is proved.

Draw the diagram for an increasing function f with n rectangles as shown in Figure (1).

Refer to Figure (1).

• The difference in upper sum and lower sum $\left(R_n-L_n\right)$ represents the sum of areas of the shaded rectangles.
• Slide the shaded rectangles to the left and stack it on the top of the leftmost shaded rectangle.
• The stacked rectangles forms a rectangle with height $f\left(b\right)-f\left(a\right)$ and width $\Delta x=\frac{b-a}{n}$.
• Hence, the area is $\frac{b-a}{n}\left[f\left(b\right)-f\left(a\right)\right]$.

Therefore, the estimate $R_n-L_n$ is reassembled and forms a single rectangle with area of $\frac{b-a}{n}\left[f\left(b\right)-f\left(a\right)\right]$.

(c)

To determine

To prove: $\left(R_n-A\right)$ is less than $\frac{b-a}{n}\left[f\left(b\right)-f\left(a\right)\right]$.

Answer to Problem 23E

Yes, the expression is proved.

Explanation of Solution

Refer to part (a).

The value of $L_n$ is less than the area A.

Therefore, it can be expressed as $A>L_n$. Also it is valid for the inequality $R_n-A<R_n-L_n$.

Substitute $\frac{b-a}{n}\left[f\left(b\right)-f\left(a\right)\right]$ for $R_n-L_n$.

$R_n-A<\frac{b-a}{n}\left[f\left(b\right)-f\left(a\right)\right]$

Therefore, the expression is valid and it is proved.