OBJECTIVES

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

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

Approximating 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.

Solution Figure 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 Figure 5(b). 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

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 Figure 6(a) 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 Figure 6(b) 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.

Figure 6

NOW WORK

Problem 5.

NEED TO REVIEW?

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

NEED TO REVIEW?

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

Approximating 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:

1. n = 2 subintervals
2. n = 5 subintervals
3. n = 10 subintervals

Solution (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 Figure 9(a). 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

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 Figure 9(b). 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 Figure 9(c). 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).

spanDEFINITIONspan 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} }}

Finding 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

Solution Figure 12 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.

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,

\begin{eqnarray*} S_{n}=\sum\limits_{i=1}^{n} f (C_i)\Delta x \underset{\underset{{\color{#0066A7}{\hbox{ \Delta x = \dfrac{10}{n}}}}}{\color{#0066A7}{\uparrow}}}{=} \sum \limits_{i=1}^{n}3x_{i}\cdot \dfrac{10}{n} \underset{\underset{{\color{#0066A7}{\hbox{ {x}_{i} = \dfrac{10i}{n}}}}}{\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

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.

Finding 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}.

Solution Figure 14 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 Figure 15 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*}

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

\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.