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)
In this section, we present a method for finding the area enclosed by the graph of afunction \(y = f(x)\) that is nonnegative on a closed interval \([a,b] \), the \(x\)-axis, and the lines \(x = a\) and \(x = b\). Thepresentation uses summation notation (\(\sum\)), which is reviewed inAppendix A.5.
The area \(A\) of a rectangle with width \(w\) and height \(l\) is given by the geometry formula \[\boxed{\bbox[#FAF8ED,5pt]{A=lw}}\]
See Figure 5.2. The graph of a constant function \(f(x)=h,\) forsome positive constant \(h,\) is a horizontal line that lies abovethe \(x\)-axis. The area enclosed by this line, the \(x\)-axis, andthe lines \(x=a\) and \(x=b\) is the rectangle whose area \(A\) isthe 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 Figure 5.3, 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\(A_{3}\).
5.1.1 Approximate the Area Under the Graph of a Function
EXAMPLE 1 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 5.4 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 thearea \(A,\) but it underestimates \(A,\) as seen in Figure 5.5(a).
Alternatively, \(A\) can be approximated by a rectangle of width \(4-2=2\) and height \(f (4) =5.\) See Figure 5.5. This approximation of thearea equals \(2\cdot 5=10\), but it overestimates \(A.\) We conclude that\[8 \lt A \lt 10\]
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The approximation of the area \(A\) can be improved by dividingthe 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 5.6 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 5.6 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 Figure 5.4 is 9 square units, obtained by using the formula for the area \(A\) ofa trapezoid with base \(b\) and parallel heights\(h_{1}\) and \(h_{2}\):\[A = \dfrac{1}{2}b (h_{1} +h_{2}) =\dfrac{1}{2}(2) (4+5) = 9.\]
Observe that as the number \(n\) of subintervals of the intervalincreases, the approximation of the area \(A\) improves. For \(n=1\), the error inapproximating \(A\) is \(1\) square unit, but for \(n=2\), the erroris 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 thelines \(x = a\) 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: \[\lbrack x_{0}, x_{1}], [x_{1}, x_{2}],\ldots ,[x_{i-1},x_{i}],\ldots , [x_{n-1}, x_{n}]\]
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each of the same length. See Figure 5.7. 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 onevery subinterval \([x_{i-1}, x_{i}]\) of \([a,b]\). By the Extreme ValueTheorem, there is a number in each subinterval where \(f\) attains itsabsolute 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 thesubinterval \([x_{i-1}, x_{i}].\) Now construct \(n\) rectangles, each having\(\Delta x\) as its base and \(f(c_{i})\) as its height, as illustrated in Figure 5.8. This produces \(n\) narrow rectangles ofuniform 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\[\begin{eqnarray*}\text{Area of the first rectangle} &=&f(c_{1})\Delta x \\\text{Area of the second rectangle} &=&f(c_{2})\Delta x \\&\vdots & \\\text{Area of the }n\text{th} \text{ (and last) rectangle} &=&f(c_{n})\Delta x\end{eqnarray*} \]
The sum \(s_{n}\) of the areas of the \(n\) rectangles approximatesthe area \(A.\) That is,\[A\approx s_{n}=f(c_{\mathbf{1}})\Delta x+f(c_{\mathbf{2}})\Delta x+~\cdots~+f(c_{i})\Delta x+~\cdots ~+f(c_{n})\Delta x=\sum\limits_{i= 1}^{n}f(c_{i})\Delta x\]
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, \(\ s_{n}\leq A.\)
NEED TO REVIEW?
Summation notation is discussed in Appendix A.5, pp. A-38 toA-43.
EXAMPLE 2 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 thatlie under the graph) for:
- \(n = 2\) subintervals
- \(n = 5\) subintervals
- \(n = 10\) 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 5.9. To compute\(s_{2}\), we need to know where \(f\) attains its minimum value ineach subinterval. Since \(f\) is an increasingfunction, the absolute minimum is attained at the left endpoint ofeach 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
347
\[s_{2}=\sum\limits_{i=1}^{2}~f(c_{i})\Delta x=\Deltax\sum\limits_{i=1}^{2}f(c_{i})\underset{\underset{\underset{\color{#0066A7}{f\left(c_{1}\right) =f\left( 0\right);f\left( c_{2}\right) =f\left(5\right)}}{\color{#0066A7}{\Delta x=5}}}{\color{#0066A7}{\uparrow}}}{=}5\,\left[ ~f(0)+f(5)\right]\underset{\underset{\underset{\color{#0066A7}{f(5)=25}}{\color{#0066A7}{f(0)=0}}}{\color{#0066A7}{\uparrow}}}{=}5(0+25)=125\]
(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 5.9. 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 5.9. The lower sum \(s_{10}\) is \[\begin{eqnarray*}s_{10} &=&\sum\limits_{i=1}^{10} f(c_{i})\Delta x=\Deltax\sum\limits_{i=1}^{10}f(c_{i})=1[ f(0)+f(1)+f(2)+\cdots +f(9) ] \\&=&0+1+4+9+16+25+36+49+64+81=285 \end{eqnarray*}\]
NOW WORK
Problem 13(a).
In general, as Figure 5.10 illustrates, the error due to using lowersums \(s_{n}\) (rectangles that lie below the graph of \(f\)) occurs because a portionof the area lies outside the rectangles. To improve the approximation of thearea, we increase the number of subintervals. For example, in Figure 5.10,there are four subintervals and the error is reduced; in Figure 5.10, thereare eight subintervals and the error is further reduced. So, by taking afiner and finer partition of the interval \([a,b]\), that is, byincreasing \(n,\) the number of subintervals, without bound, we can make the sum of theareas of the rectangles as close as we please to the actual area. (A proof ofthis statement is usually found in books on advanced calculus.)
In Section 5.3, we see that, for functions that are continuous on a closedinterval, \(\lim\limits_{n\rightarrow \infty }s_{n}\) always exists. With thisdiscussion in mind, we now define the area under the graph of afunction \(f\) from \(a\) to \(b\).
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DEFINITION Area \(A\) Under the Graph of a Function from \(a\) to \(b\)
Suppose a function \(f\) is nonnegative and continuous on a closedinterval \([a,b] \). 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 \(f (c_{i}) \) equalthe absolute minimum value of \(f\) on this subinterval. Form the lower sums \[s_{n}=\sum\limits_{i= 1}^{n}f({\rm c_{i}})\Delta x=f({\rm c}_{\mathbf{1}})\Deltax+ \cdots +f(c_{n})\Delta x\]
The area \(A\) under the graph of \(f\) from \(a\) to \(b\) is the number\[\boxed{\bbox[5pt] {A=\lim\limits_{n\rightarrow \infty }s_{n} }}\]
The area \(A\) is defined using lower sums \(s_{n}\) (rectangles that lie belowthe 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 valueof \(f\) on the \(i\)th subinterval, as shown in Figure 5.11. Thecorresponding upper sums \(S_{n}\) (rectangles that lie above the graph of\(f\)) overestimates the area \(A\). So, \(S_{n}\geq A\). Itcan be shown that as \(n\) increases without bound, the limit of theupper sums \(S_{n}\) equals the limit of the lower sums \(s_{n}.\) That is,\[\boxed{\bbox[#FAF8ED,5pt]{\lim\limits_{n\rightarrow\infty}s_{n}=\lim\limits_{n\rightarrow \infty }S_{n}=A }}\]
5.1.2 Find the Area Under the Graph of a Function
In the next example, instead of using a specific number ofrectangles to approximate area, we partition the interval\([a,b]\) into \(n\) subintervals, obtaining \(n\) rectangles. Byletting \(n\rightarrow \infty ,\) we find the actual area under the graph of \(f\) from \(a\) to \(b.\)
EXAMPLE 3 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 \(A = \lim\limits_{n\rightarrow \infty }S_{n}\).
Solution Figure 5.12 illustrates the area \(A\). We partition theclosed 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 interms 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 5.13.
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To find \(A\) using upper sums \(S_{n}\) (rectangles that lie above the graphof \(f)\), we need the absolute maximum value of \(f\) in each subinterval.Since \(f(x)=3x\) is an increasing function, the absolute maximum occurs atthe right endpoint \(x_{i}= i\left(\dfrac{10}{n}\right) \) of eachsubinterval. 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*}\]
NEED TO REVIEW?
Summation properties are discussed in Appendix A.5, pp. A-38 toA-43.
RECALL
\(\sum\limits_{i=1}^{n}i=\dfrac{n(n+1) }{2}\)
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.
NOTE
The area found in Example 3 is that of a triangle. So, we canverify that \(A=150\) by using the formula for the area \(A\) of atriangle with base \(b\) and height \(h\): \[A=\dfrac{1}{2}bh = \dfrac{1}{2} (10)(30) = 150 \]
EXAMPLE 4 Finding Area Using Lower Sums
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 liebelow the graph of \(f\)). Then \(A=\lim\limits_{n\rightarrow \infty}{}s_{n}\).
Solution Figure 5.14 shows the area under the graphof \(f\) and a typical rectangle that lies below the graph. We partition the closedinterval \([0,4]\) 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 5.15 on page 350 illustrates, the endpoints ofeach subinterval, written in terms of \(n\), are\[\begin{eqnarray*}x_{0} &=& 0,\ x_{1}=1\left( \dfrac{4}{n}\right)\!,\ x_{2}=2\left( {{\dfrac{{4}}{{n}}}}\right), \ldots , x_{i-1}=(i-1)\left( {{\dfrac{{4}}{{n}}}}\right), \\x_{i} &=& i\left( {{\dfrac{{4}}{{n}}}}\right)\!, \ldots , x_{n}=n\left( {{\dfrac{{4}}{{n}}}}\right) =4\end{eqnarray*}\]
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To find \(A\) using lower sums \(s_{n}\) (rectangles that lie below the graphof \(f\)), we must find the absolute minimum value of \(f\) on each subinterval.Since the function \(f\) is a decreasing function, the absolute minimum occursat 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\[\begin{eqnarray*}s_{n} &=&\sum\limits_{i=1}^{n}f(c_{i})\Delta x = \sum\limits_{i=1}^{n}\left[ {16-{{\left( {{\dfrac{{4i}}{{n}}}}\right) }}^{2}} \right] \left( {{\dfrac{{4}}{{n}}}}\right){\color{#0066A7}{(c_i = \dfrac{4i}{n}); f(c_{i}) = 16 - c_{i}^{2}}}\\&=& \sum\limits_{i=1}^{n}\left[ \dfrac{64}{n}-\dfrac{64i^{2}}{n^{3}}\right] \\&=& \dfrac{64}{n}{{{\sum\limits_{i=1}^{n}1}}}-\dfrac{64}{n^{3}}\sum\limits_{i=1}^{n}i^{2} \\&=& \dfrac{64}{n}(n) -\dfrac{64}{n^{3}}\left[ \dfrac{n(n+1) (2n+1) }{6}\right] = 64-\dfrac{32}{3n^{2}}\left[ 2n^{2}+3n+1\right] \\&=& 64-\dfrac{64}{3}-\dfrac{32}{n}-\dfrac{32}{3n^{2}}=\dfrac{128}{3}-\dfrac{32}{n}-\dfrac{32}{3n^{2}} \end{eqnarray*}\]
RECALL
\[\sum\limits_{i=1}^{n} 1=n;\]\[\sum\limits_{i=1}^{n} {i}^{2} = \dfrac{{n}({n+1})(2n+1)}{6}\]
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 findareas 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 5.5, we present the Fundamental Theorem of Calculus, which provides arelatively simple way to find area.