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.