sullivancalculus

5.1 Area

OBJECTIVES

When you finish this section, you should be able to:

Approximate the area under the graph of a function (p. 344)
Find the area under the graph of a function (p. 348)

Figure 2 $A=h(b-a)$ .

In this section, we present a method for finding the area enclosed by the graph of a function $y = f(x)$ that is nonnegative on a closed interval $[a,b]$ , the $x$ -axis, and the lines $x = a$ and $x = b$ . The presentation uses summation notation ( $\sum$ ), which is reviewed in Appendix A.5.

The area $A$ of a rectangle with width $w$ and height $l$ is given by the geometry formula $\boxed{\bbox[#FAF8ED,5pt]{A=lw}}$

Figure 3 $A=A_1+A_2+A_3$

See . The graph of a constant function $f(x)=h,$ for some positive constant $h,$ is a horizontal line that lies above the $x$ -axis. The area enclosed by this line, the $x$ -axis, and the lines $x=a$ and $x=b$ is the rectangle whose area $A$ is the product of the width $(b-a)$ and the height $h$ . $\boxed{\bbox[#FAF8ED,5pt]{A= h (b-a)}}$

If the graph of $y = f(x)$ consists of three horizontal lines, each of positive height as shown in , the area $A$ enclosed by the graph of $f$ , the $x$ -axis, and the lines $x = a$ and $x = b$ is the sum of the rectangular areas $A_{1}$ , $A_{2}$ , and $A_{3}$ .

Approximate the Area Under the Graph of a Function

EXAMPLE 1Approximating the Area Under the Graph of a Function

Approximate the area $A$ enclosed by the graph of $f(x) =\dfrac{1}{2}x+3,$ the $x$ -axis, and the lines $x=2$ and $x=4$ .

illustrates the area $A$ to be approximated.

We begin by drawing a rectangle of width $4-2=2$ and height $f (2)=4.$ The area of the rectangle, $2\cdot 4=8,$ approximates the area $A,$ but it underestimates $A,$ as seen in Figure 5(a).

Alternatively, $A$ can be approximated by a rectangle of width $4-2=2$ and height $f (4) =5.$ See . This approximation of the area equals $2\cdot 5=10$ , but it overestimates $A.$ We conclude that $8 \lt A \lt 10$

Figure 4

Figure 5

345

The approximation of the area $A$ can be improved by dividing the closed interval $[2, 4]$ into two subintervals, $[2, 3]$ and $[3, 4].$ Now we draw two rectangles: one rectangle with width $3-2=1$ and height $f(2) =\dfrac{1}{2}\cdot 2+3=4$ ; the other rectangle with width $4-3=1$ and height $f (3) =\dfrac{1}{2} \cdot 3+3=\dfrac{9}{2}.$ As illustrates, the sum of the areas of the two rectangles $1\cdot 4+1\cdot \dfrac{9}{2}=\dfrac{17}{2}=8.5$ underestimates the area.

Now we repeat this process by drawing two rectangles,one of width $1$ and height $f(3) =\dfrac{9}{2};$ the other of width $1$ and height $f(4) =\dfrac{1}{2}\cdot 4+3=5$ . As illustrates, the sum of the areas of these two rectangles, $1\cdot \dfrac{9}{2}+1\cdot 5=\dfrac{19}{2}=9.5$

overestimates the area. We conclude that $8.5\lt A\lt9.5$

obtaining a better approximation to the area.

NOTE

The actual area in is 9 square units, obtained by using the formula for the area $A$ of a trapezoid with base $b$ and parallel heights $h_{1}$ and $h_{2}$ : $A = \dfrac{1}{2}b (h_{1} +h_{2}) =\dfrac{1}{2}(2) (4+5) = 9.$

Figure 6

Observe that as the number $n$ of subintervals of the interval increases, the approximation of the area $A$ improves. For $n=1$ , the error in approximating $A$ is $1$ square unit, but for $n=2$ , the error is only $0.5$ square unit.

NOW WORK

Problem 5.

In general, the procedure for approximating the area $A$ is based on the idea of summing the areas of rectangles. We shall refer to the area $A$ enclosed by the graph of a function $y=f(x)\geq 0$ , the $x$ -axis, and the lines $x = a$ and $x = b$ as the area under the graph of $\boldsymbol{f}$ from $\boldsymbol{a}$ to $\boldsymbol{b}$ .

We make two assumptions about the function $f$ :

$f$ is continuous on the closed interval $[a,b]$ .
$f$ is nonnegative on the closed interval $[a,b]$ .

We divide, or partition, the interval $[a,b]$ into $n$ nonoverlapping subintervals: $\lbrack x_{0}, x_{1}], [x_{1}, x_{2}],\ldots , [x_{i-1},x_{i}],\ldots , [x_{n-1}, x_{n}]$

346

Figure 7

each of the same length. See . Since there are $n$ subintervals and the length of the interval $[a,b]$ is $b-a,$ the common length $\Delta x$ of each subinterval is $\boxed{\bbox[#FAF8ED,5pt]{\Delta x=\dfrac{b-a}{n}}}$

NEED TO REVIEW?

The Extreme Value Theorem is discussed in Section 4.2, pp. xx-xx.

Since $f$ is continuous on the closed interval $[a,b]$ , it is continuous on every subinterval $[x_{i-1}, x_{i}]$ of $[a,b]$ . By the Extreme Value Theorem, there is a number in each subinterval where $f$ attains its absolute minimum. Label these numbers $c_{1},$ $c_{2},$ $c_{3},\ldots , c_{n}$ , so that $f(c_{i})$ is the absolute minimum value of $f$ in the subinterval $[x_{i-1}, x_{i}].$ Now construct $n$ rectangles, each having $\Delta x$ as its base and $f(c_{i})$ as its height, as illustrated in . This produces $n$ narrow rectangles of uniform width $\Delta x=\dfrac{b-a}{n}$ and heights $f(c_{1}) , f(c_{2}), \ldots, f(c_{n}) ,$ respectively. The areas of the $n$ rectangles are

Figure 8 $f(c_i)$ is the absolute minimum value of $f$ on $[x_{i-1}, x_i]$ .

$\begin{eqnarray*} \text{Area of the first rectangle} &=&f(c_{1})\Delta x \\ \text{Area of the second rectangle} &=&f(c_{2})\Delta x \\ &\vdots & \\ \text{Area of the }n\text{th} \text{ (and last) rectangle} &=&f(c_{n})\Delta x \end{eqnarray*}$

The sum $s_{n}$ of the areas of the $n$ rectangles approximates the area $A.$ That is,

NEED TO REVIEW?

Summation notation is discussed in Appendix A.5, pp. A-38 to A-43.

$A\approx s_{n}=f(c_{\mathbf{1}})\Delta x+f(c_{\mathbf{2}})\Delta x+~\cdots ~+f(c_{i})\Delta x+~\cdots ~+f(c_{n})\Delta x=\sum\limits_{i= 1}^{n}f(c_{i})\Delta x$

Since the rectangles used to approximate the area $A$ lie under the graph of $f,$ the sum $s_{n},$ called a lower sum, underestimates $A$ . That is, $\ s_{n}\leq A.$

EXAMPLE 2Approximating Area Using Lower Sums

Approximate the area $A$ under the graph of $f(x) = x^{2}$ from $0$ to $10$ by using lower sums $s_{n}$ (rectangles that lie under the graph) for:

(a) subintervals

(b) subintervals

$n = 2$ $n = 5$ $n = 10$

(a) For $n=2$ , we partition the closed interval $[0,10]$ into two subintervals $[0,5]$ and $[5,10]$ , each of length $\Delta x=\dfrac{10-0}{2}=5$ . See . To compute $s_{2}$ , we need to know where $f$ attains its minimum value in each subinterval. Since $f$ is an increasing function, the absolute minimum is attained at the left endpoint of each subinterval. So, for $n=2,$ the minimum of $f$ on $[0,5]$ occurs at $0$ and the minimum of $f$ on $[5,10]$ occurs at 5. The lower sum $s_{2}$ is

Figure 9

347

$s_{2}=\sum\limits_{i=1}^{2}~f(c_{i})\Delta x=\Delta x\sum\limits_{i=1}^{2}f(c_{i}) \underset{\underset{\underset{\color{#0066A7}{f\left(c_{1}\right) =f\left( 0\right);f\left( c_{2}\right) =f\left(5\right)}}{\color{#0066A7}{\Delta x=5}}}{\color{#0066A7}{\uparrow}}}{=}5\,\left[ ~f(0)+f(5)\right] \underset{\underset{\underset{\color{#0066A7}{f(5)=25}}{\color{#0066A7}{f(0)=0}}}{\color{#0066A7}{\uparrow}}}{=}5(0+25)=125$

(b) For $n=5$ , partition the interval $[0,10]$ into five subintervals $[0,2] ,$ $[2,4] ,$ $[4,6] ,$ $[6,8] ,$ $[8,10]$ , each of length $\Delta x = \dfrac{10-0}{5}=2$ . See . The lower sum $s_{5}$ is $\begin{eqnarray*} s_{5} &=&\sum\limits_{i=1}^{5} f(c_{i})\Delta x=\Delta x\sum\limits_{i=1}^{5}f(c_{i})=2 [ f(0)+f(2)+f(4) +f(6)+f(8)] \\ &=&2(0+4+16+36+64)=240 \end{eqnarray*}$

(c) For $n=10$ , partition $[0,10]$ into $10$ subintervals, each of length $\Delta x=\dfrac{10-0}{10}=1$ . See . The lower sum $s_{10}$ is $\begin{eqnarray*} s_{10} &=&\sum\limits_{i=1}^{10} f(c_{i})\Delta x=\Delta x\sum\limits_{i=1}^{10}f(c_{i})=1[ f(0)+f(1)+f(2)+\cdots +f(9) ] \\ &=&0+1+4+9+16+25+36+49+64+81=285 \end{eqnarray*}$

NOW WORK

Problem 13(a).

In general, as illustrates, the error due to using lower sums $s_{n}$ (rectangles that lie below the graph of $f$ ) occurs because a portion of the area lies outside the rectangles. To improve the approximation of the area, we increase the number of subintervals. For example, in , there are four subintervals and the error is reduced; in , there are eight subintervals and the error is further reduced. So, by taking a finer and finer partition of the interval $[a,b]$ , that is, by increasing $n,$ the number of subintervals, without bound, we can make the sum of the areas of the rectangles as close as we please to the actual area. (A proof of this statement is usually found in books on advanced calculus.)

Figure 10

In , we see that, for functions that are continuous on a closed interval, $\lim\limits_{n\rightarrow \infty }s_{n}$ always exists. With this discussion in mind, we now define the area under the graph of a function $f$ from $a$ to $b$ .

348

Area $A$ Under the Graph of a Function from $a$ to $b$

Suppose a function $f$ is nonnegative and continuous on a closed interval $[a,b]$ . Partition $[a,b]$ into $n$ subintervals $[ x_{0},x_{1}] , [ x_{1},x_{2}] , \ldots , [ x_{i-1},x_{i}] , \ldots , [ x_{n-1},x_{n}] ,$ each of length $\Delta x=\dfrac{b-a}{n}$

In each subinterval $[ x_{i-1},x_{i}] ,$ let $f (c_{i})$ equal the absolute minimum value of $f$ on this subinterval. Form the lower sums $s_{n}=\sum\limits_{i= 1}^{n}f({\rm c_{i}})\Delta x=f({\rm c}_{\mathbf{1}})\Delta x+ \cdots +f(c_{n})\Delta x$

The area $A$ under the graph of $f$ from $a$ to $b$ is the number $\boxed{\bbox[5pt] {A=\lim\limits_{n\rightarrow \infty }s_{n} }}$

Figure 11 $f(C_i)$ is the absolute maximum value of $f$ on $[x_{i-1}, x_i]$ .

The area $A$ is defined using lower sums $s_{n}$ (rectangles that lie below the graph of $f$ ). By a parallel argument, we can choose values $C_{1}, C_{2}, \ldots , C_{n}$ so that the height $f (C_{i})$ of the $i$ th rectangle is the absolute maximum value of $f$ on the $i$ th subinterval, as shown in . The corresponding upper sums $S_{n}$ (rectangles that lie above the graph of $f$ ) overestimates the area $A$ . So, $S_{n}\geq A$ . It can be shown that as $n$ increases without bound, the limit of the upper sums $S_{n}$ equals the limit of the lower sums $s_{n}.$ That is, $\boxed{\bbox[#FAF8ED,5pt]{\lim\limits_{n\rightarrow \infty}s_{n}=\lim\limits_{n\rightarrow \infty }S_{n}=A }}$

Find the Area Under the Graph of a Function

In the next example, instead of using a specific number of rectangles to approximate area, we partition the interval $[a,b]$ into $n$ subintervals, obtaining $n$ rectangles. By letting $n\rightarrow \infty ,$ we find the actual area under the graph of $f$ from $a$ to $b.$

EXAMPLE 3Finding Area Using Upper Sums

Find the area $A$ under the graph of $f(x) = 3x$ from $0$ to $10$ using upper sums $S_{n}$ (rectangles that lie above the graph of $f)$ . Then $A = \lim\limits_{n\rightarrow \infty }S_{n}$ .

Figure 12 $f(x) = 3x, 0\leq x \leq 10$

illustrates the area $A$ . We partition the closed interval $[0,10]$ into $n$ subintervals ${[{x_{0},x_{1}}]}, {[{x_{1},x_{2}}]}, \ldots, [x_{i-1},x_{i}] , \ldots , {{[{x_{n-1},x_{n}}]}}$

where $0=x_{0}\lt x_{1}\lt x_{2}\lt \cdots \lt x_{i}\lt \cdots \lt x_{n-1}\lt x_{n}=10$

and each subinterval is of length $\Delta x = {\dfrac{{10 - 0}}{{n}}} = {\dfrac{{10}}{{n}}}$

The coordinates of the endpoints of each subinterval, written in terms of $n,$ are $\begin{eqnarray*} x_{0} &=& 0,\ x_{1}={\dfrac{{10}}{{n}}},\ x_{2}=2\left( {{\dfrac{{10}}{{n}}}}\right), \ldots , x_{i-1}=(i-1) \left( {{\dfrac{{10}}{{n}}}}\right), \\ x_{i} &=& i\left( {{\dfrac{{10}}{{n}}}}\right), \ldots , x_{n}=n\left( {{\dfrac{{10}}{{n}}}}\right) =10 \end{eqnarray*}$

as illustrated in Figure 13.

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Figure 13

To find $A$ using upper sums $S_{n}$ (rectangles that lie above the graph of $f)$ , we need the absolute maximum value of $f$ in each subinterval. Since $f(x)=3x$ is an increasing function, the absolute maximum occurs at the right endpoint $x_{i}= i\left(\dfrac{10}{n}\right)$ of each subinterval. So,

NEED TO REVIEW?

Summation properties are discussed in Appendix A.5, pp. A-38 to A-43.

$\begin{eqnarray*} S_{n}=\sum\limits_{i=1}^{n} f (C_i)\Delta x \underset{\underset{{\color{#0066A7}{\hbox{<span data-block_type="tex"> \Delta x = \dfrac{10}{n}</span>}}}}{\color{#0066A7}{\uparrow}}}{=} \sum \limits_{i=1}^{n}3x_{i}\cdot \dfrac{10}{n} \underset{\underset{{\color{#0066A7}{\hbox{<span data-block_type="tex"> {x}_{i} = \dfrac{10i}{n}</span>}}}}{\color{#0066A7}{\uparrow}}}{=} {\sum\limits_{i=1}^{n}}\left( 3\cdot \dfrac{10i}{n}\right) \left( \dfrac{10}{n}\right) = \sum\limits_{i=1}^{n}\dfrac{300}{n^2}i \end{eqnarray*}$

Using summation properties, we get

RECALL

$\sum\limits_{i=1}^{n}i=\dfrac{n(n+1) }{2}$

$S_{n}= \sum\limits_{i=1}^{n} \dfrac{300}{n^2} i =\dfrac{300}{n^2} \sum\limits_{i=1}^{n} i = \dfrac{300}{n^2}\ \dfrac{n(n+1)}{2} = 150\left(\dfrac{n+1}{n}\right) =150\left(1+\dfrac{1}{n}\right)$

Then $A=\lim\limits_{n\rightarrow \infty }S_{n}=\lim\limits_{n\rightarrow \infty } \left[150 \left( 1+\dfrac{1}{n} \right) \right] =150\lim\limits_{n\rightarrow \infty }\left(1+\dfrac{1}{n}\right) = 150$

The area $A$ under the graph of $f(x)=3x$ from $0$ to $10$ is $150$ square units.

NOTE The area found in Example 3 is that of a triangle. So, we can verify that $A=150$ by using the formula for the area $A$ of a triangle with base $b$ and height $h$ : $A=\dfrac{1}{2}bh = \dfrac{1}{2} (10)(30) = 150$

EXAMPLE 4Finding Area Using Lower Sums

Figure 14 $f(x) = 16-x^2, 0 \leq x \leq 4$

Find the area $A$ under the graph of $f(x) = 16 - x^{2}$ from $0$ to $4$ by using lower sums $s_{n}$ (rectangles that lie below the graph of $f$ ). Then $A=\lim\limits_{n\rightarrow \infty }{}s_{n}$ .

shows the area under the graph of $f$ and a typical rectangle that lies below the graph. We partition the closed interval $[0,4]$ into $n$ subintervals ${[{x_{0},x_{1}}]}, {[{x_{1},x_{2}}]}, \ldots , [x_{i-1},x_{i}], \ldots , {[{x_{n-1},x_{n}}]}$

where $0=x_{0}\lt x_{1}\lt\cdots \lt x_{i}\lt\cdots \lt x_{n-1}\lt x_{n}=4$

and each interval is of length $\Delta x = \dfrac{{4-0}}{n} = \dfrac{4}{n}$

As on page 350 illustrates, the endpoints of each subinterval, written in terms of $n$ , are $\begin{eqnarray*} x_{0} &=& 0,\ x_{1}=1\left( \dfrac{4}{n}\right)\!,\ x_{2}=2\left( {{\dfrac{{4}}{{n}}}}\right), \ldots , x_{i-1}=(i-1) \left( {{\dfrac{{4}}{{n}}}}\right), \\ x_{i} &=& i\left( {{\dfrac{{4}}{{n}}}} \right)\!, \ldots , x_{n}=n\left( {{\dfrac{{4}}{{n}}}}\right) =4 \end{eqnarray*}$

350

Figure 15

To find $A$ using lower sums $s_{n}$ (rectangles that lie below the graph of $f$ ), we must find the absolute minimum value of $f$ on each subinterval. Since the function $f$ is a decreasing function, the absolute minimum occurs at the right endpoint of each subinterval. So, $s_{n}={\sum\limits_{i=1}^{n}} f (c_{i})\Delta x$

Since $c_{i}=i \left( \dfrac{4}{n}\right) =\dfrac{4i}{n}$ and $\Delta x=\dfrac{4}{n}$ , we have

RECALL

$\sum\limits_{i=1}^{n} 1=n;$ $\sum\limits_{i=1}^{n} {i}^{2} = \dfrac{{n}({n+1}) (2n+1)}{6}$

$\begin{eqnarray*} s_{n} &=&\sum\limits_{i=1}^{n}f(c_{i})\Delta x = \sum \limits_{i=1}^{n}\left[ {16-{{\left( {{\dfrac{{4i}}{{n}}}}\right) }}^{2}} \right] \left( {{\dfrac{{4}}{{n}}}}\right) {\color{#0066A7}{(c_i = \dfrac{4i}{n}); f(c_{i}) = 16 - c_{i}^{2}}}\\ &=& \sum\limits_{i=1}^{n}\left[ \dfrac{64}{n}-\dfrac{64i^{2}}{n^{3}}\right] \\ &=& \dfrac{64}{n}{{{\sum\limits_{i=1}^{n}1}}}-\dfrac{64}{n^{3}}\sum\limits_{i=1}^{n}i^{2} \\ &=& \dfrac{64}{n}(n) -\dfrac{64}{n^{3}}\left[ \dfrac{n(n+1) (2n+1) }{6}\right] = 64-\dfrac{32}{3n^{2}}\left[ 2n^{2}+3n+1\right] \\ &=& 64-\dfrac{64}{3}-\dfrac{32}{n}-\dfrac{32}{3n^{2}}=\dfrac{128}{3}-\dfrac{32}{n}-\dfrac{32}{3n^{2}} \end{eqnarray*}$

Then $A=\lim\limits_{n\rightarrow \infty }s_{n}={\lim\limits_{n\rightarrow \infty }}\left( {{\dfrac{{128}}{{3}}}}-{{\dfrac{{32}}{{n}}}-{\dfrac{{32}}{{3n^{2}}}}}\right) ={\dfrac{{128}}{{3}}}$

The area $A$ under the graph of $f (x) =16-x^{2}$ from $0$ to $4$ is $\dfrac{128}{3}$ square units.

NOW WORK

Problem 29.

The previous two examples illustrate just how complex it can be to find areas using lower sums and/or upper sums. In the next section, we define an integral and show how it can be used to find area. Then in Section 5.3, we present the Fundamental Theorem of Calculus, which provides a relatively simple way to find area.

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5.1 Area

OBJECTIVES

EXAMPLE 1Approximating the Area Under the Graph of a Function

NOTE

NOW WORK

NEED TO REVIEW?

NEED TO REVIEW?

EXAMPLE 2Approximating Area Using Lower Sums

NOW WORK

DEFINITION Area A Under the Graph of a Function from a to b

EXAMPLE 3Finding Area Using Upper Sums

NEED TO REVIEW?

RECALL

EXAMPLE 4Finding Area Using Lower Sums

RECALL

NOW WORK

Area $A$ Under the Graph of a Function from $a$ to $b$