Approximating Areas

Archimedes was fascinated with calculating the areas of various shapes—in other words, the amount of space enclosed by the shape. He used a process that has come to be known as the method of exhaustion, which used smaller and smaller shapes, the areas of which could be calculated exactly, to fill an irregular region and thereby obtain closer and closer approximations to the total area. In this process, an area bounded by curves is filled with rectangles, triangles, and shapes with exact area formulas. These areas are then summed to approximate the area of the curved region.

In this section, we develop techniques to approximate the area between a curve, defined by a function

f (x),

Like Archimedes, we first approximate the area under the curve using shapes of known area (namely, rectangles). By using smaller and smaller rectangles, we get closer and closer approximations to the area. Taking a limit allows us to calculate the exact area under the curve.

Let’s start by introducing some notation to make the calculations easier. We then consider the case when

f (x)

is continuous and nonnegative. Later in the chapter, we relax some of these restrictions and develop techniques that apply in more general cases.

Sigma (Summation) Notation

As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. This process often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look at some new notation here, called sigma notation (also known as summation notation). The Greek capital letter

Σ,

sigma, is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20 without sigma notation, we have to write

which is better, but still cumbersome. With sigma notation, we write this sum as

describes the terms to be added, and the i is called the index. Each term is evaluated, then we sum all the values, beginning with the value when

i = 1

Note that the index is used only to keep track of the terms to be added; it does not factor into the calculation of the sum itself. The index is therefore called a dummy variable. We can use any letter we like for the index. Typically, mathematicians use i, j, k, m, and n for indices.

The properties associated with the summation process are given in the following rule.

Proof

We prove properties 2. and 3. here, and leave proof of the other properties to the Exercises.

A few more formulas for frequently found functions simplify the summation process further. These are shown in the next rule, for sums and powers of integers, and we use them in the next set of examples.

Approximating Area

Now that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let

f (x)

How do we approximate the area under this curve? The approach is a geometric one. By dividing a region into many small shapes that have known area formulas, we can sum these areas and obtain a reasonable estimate of the true area. We begin by dividing the interval

[a, b]

We denote the width of each subinterval with the notation Δx, so

Δ x = \frac{b - a}{n}

into subintervals by selecting points from within the interval is used quite often in approximating the area under a curve, so let’s define some relevant terminology.

We can use this regular partition as the basis of a method for estimating the area under the curve. We next examine two methods: the left-endpoint approximation and the right-endpoint approximation.

The second method for approximating area under a curve is the right-endpoint approximation. It is almost the same as the left-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of each subinterval.

We then form six rectangles by drawing vertical lines perpendicular to

x_{i - 1},

the left endpoint of each subinterval. We determine the height of each rectangle by calculating

f (x_{i - 1})

The intervals are

[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2], [2, 2.5], [2.5, 3] .

We find the area of each rectangle by multiplying the height by the width. Then, the sum of the rectangular areas approximates the area between

f (x)

and the x-axis. When the left endpoints are used to calculate height, we have a left-endpoint approximation. Thus,

by Δx to find the rectangular areas, and then add them. This is a right-endpoint approximation of the area under

f (x) .

Looking at [link] and the graphs in [link], we can see that when we use a small number of intervals, neither the left-endpoint approximation nor the right-endpoint approximation is a particularly accurate estimate of the area under the curve. However, it seems logical that if we increase the number of points in our partition, our estimate of A will improve. We will have more rectangles, but each rectangle will be thinner, so we will be able to fit the rectangles to the curve more precisely.

We can demonstrate the improved approximation obtained through smaller intervals with an example. Let’s explore the idea of increasing n, first in a left-endpoint approximation with four rectangles, then eight rectangles, and finally 32 rectangles. Then, let’s do the same thing in a right-endpoint approximation, using the same sets of intervals, of the same curved region. [link] shows the area of the region under the curve

f (x) = {(x - 1)}^{3} + 4

[link] shows the same curve divided into eight subintervals. Comparing the graph with four rectangles in [link] with this graph with eight rectangles, we can see there appears to be less white space under the curve when

n = 8 .

This white space is area under the curve we are unable to include using our approximation. The area of the rectangles is

The graph in [link] shows the same function with 32 rectangles inscribed under the curve. There appears to be little white space left. The area occupied by the rectangles is

We can carry out a similar process for the right-endpoint approximation method. A right-endpoint approximation of the same curve, using four rectangles ([link]), yields an area

Based on these figures and calculations, it appears we are on the right track; the rectangles appear to approximate the area under the curve better as n gets larger. Furthermore, as n increases, both the left-endpoint and right-endpoint approximations appear to approach an area of 8 square units. [link] shows a numerical comparison of the left- and right-endpoint methods. The idea that the approximations of the area under the curve get better and better as n gets larger and larger is very important, and we now explore this idea in more detail.

Forming Riemann Sums

Converging Values of Left- and Right-Endpoint Approximations as n Increases
Values of n	Approximate Area L_n	Approximate Area R_n
$n = 4$	7.5	8.5
$n = 8$	7.75	8.25
$n = 32$	7.94	8.06

So far we have been using rectangles to approximate the area under a curve. The heights of these rectangles have been determined by evaluating the function at either the right or left endpoints of the subinterval

[x_{i - 1}, x_{i}] .

In reality, there is no reason to restrict evaluation of the function to one of these two points only. We could evaluate the function at any point c_i in the subinterval

[x_{i - 1}, x_{i}],

as the height of our rectangle. This gives us an estimate for the area of the form

A sum of this form is called a Riemann sum, named for the 19th-century mathematician Bernhard Riemann, who developed the idea.

Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as n get larger and larger. The same thing happens with Riemann sums. Riemann sums give better approximations for larger values of n. We are now ready to define the area under a curve in terms of Riemann sums.

Some subtleties here are worth discussing. First, note that taking the limit of a sum is a little different from taking the limit of a function

f (x)

as x goes to infinity. Limits of sums are discussed in detail in the chapter on Sequences and Series; however, for now we can assume that the computational techniques we used to compute limits of functions can also be used to calculate limits of sums.

Second, we must consider what to do if the expression converges to different limits for different choices of

{x_{i}^{*}} .

Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if

f (x)

exists and is unique (in other words, it does not depend on the choice of

{x_{i}^{*}}).

We look at some examples shortly. But, before we do, let’s take a moment and talk about some specific choices for

{x_{i}^{*}} .

gives us an estimate of the area under the curve, we don’t necessarily know whether that estimate is too high (overestimate) or too low (underestimate). If it is important to know whether our estimate is high or low, we can select our value for

{x_{i}^{*}}

is called an upper sum. Similarly, if we want an underestimate, we can choose

{x_{i}^{*}}

In this case, the associated Riemann sum is called a lower sum. Note that if

f (x)

then the maximum and minimum values of the function occur at the endpoints of the subintervals, so the upper and lower sums are just the same as the left- and right-endpoint approximations.

Key Concepts

Key Equations

State whether the given sums are equal or unequal.

$\sum_{i = 1}^{10} i$
and
$\sum_{k = 1}^{10} k$
$\sum_{i = 1}^{10} i$
and
$\sum_{i = 6}^{15} (i - 5)$
$\sum_{i = 1}^{10} i (i - 1)$
and
$\sum_{j = 0}^{9} (j + 1) j$
$\sum_{i = 1}^{10} i (i - 1)$
and
$\sum_{k = 1}^{10} (k^{2} - k)$

a. They are equal; both represent the sum of the first 10 whole numbers. b. They are equal; both represent the sum of the first 10 whole numbers. c. They are equal by substituting $j = i - 1 .$

d. They are equal; the first sum factors the terms of the second.

In the following exercises, use the rules for sums of powers of integers to compute the sums.

\sum_{i = 5}^{10} i

\sum_{i = 5}^{10} i^{2}

385 - 30 = 355

Suppose that $\sum_{i = 1}^{100} a_{i} = 15$

and $\sum_{i = 1}^{100} b_{i} = −12 .$

In the following exercises, compute the sums.

\sum_{i = 1}^{100} (a_{i} + b_{i})

\sum_{i = 1}^{100} (a_{i} - b_{i})

15 - (−12) = 27

\sum_{i = 1}^{100} (3 a_{i} - 4 b_{i})

\sum_{i = 1}^{100} (5 a_{i} + 4 b_{i})

5 (15) + 4 (−12) = 27

In the following exercises, use summation properties and formulas to rewrite and evaluate the sums.

\sum_{k = 1}^{20} 100 (k^{2} - 5 k + 1)

\sum_{j = 1}^{50} (j^{2} - 2 j)

\sum_{j = 1}^{50} j^{2} - 2 \sum_{j = 1}^{50} j = \frac{(50) (51) (101)}{6} - \frac{2 (50) (51)}{2} = 40, ​ 375

\sum_{j = 11}^{20} (j^{2} - 10 j)

\sum_{k = 1}^{25} [{(2 k)}^{2} - 100 k]

4 \sum_{k = 1}^{25} k^{2} - 100 \sum_{k = 1}^{25} k = \frac{4 (25) (26) (51)}{9} - 50 (25) (26) = −10, ​ 400

Let $L_{n}$

denote the left-endpoint sum using n subintervals and let $R_{n}$

denote the corresponding right-endpoint sum. In the following exercises, compute the indicated left and right sums for the given functions on the indicated interval.

L₄ for $f (x) = \frac{1}{x - 1}$

on $[2, 3]$

R₄ for $g (x) = cos (π x)$

on $[0, 1]$

R_{4} = 0.25

L₆ for $f (x) = \frac{1}{x (x - 1)}$

on $[2, 5]$

R₆ for $f (x) = \frac{1}{x (x - 1)}$

on $[2, 5]$

R_{6} = 0.372

R₄ for $\frac{1}{x^{2} + 1}$

on $[−2, 2]$

L₄ for $\frac{1}{x^{2} + 1}$

on $[−2, 2]$

L_{4} = 2.20

R₄ for $x^{2} - 2 x + 1$

on $[0, 2]$

L₈ for $x^{2} - 2 x + 1$

on $[0, 2]$

L_{8} = 0.6875

Compute the left and right Riemann sums—L₄ and R₄, respectively—for $f (x) = (2 - \| x \|)$

on $[−2, 2] .$

Compute their average value and compare it with the area under the graph of f.

Compute the left and right Riemann sums—L₆ and R₆, respectively—for $f (x) = (3 - \| 3 - x \|)$

on $[0, 6] .$

Compute their average value and compare it with the area under the graph of f.

L_{6} = 9.000 = R_{6} .

The graph of f is a triangle with area 9.

Compute the left and right Riemann sums—L₄ and R₄, respectively—for $f (x) = \sqrt{4 - x^{2}}$

on $[−2, 2]$

and compare their values.

Compute the left and right Riemann sums—L₆ and R₆, respectively—for $f (x) = \sqrt{9 - {(x - 3)}^{2}}$

on $[0, 6]$

and compare their values.

L_{6} = 13.12899 = R_{6} .

They are equal.

Express the following endpoint sums in sigma notation but do not evaluate them.

L₃₀ for $f (x) = x^{2}$

on $[1, 2]$

L₁₀ for $f (x) = \sqrt{4 - x^{2}}$

on $[−2, 2]$

L_{10} = \frac{4}{10} \sum_{i = 1}^{10} \sqrt{4 - (−2 + 4 \frac{(i - 1)}{10})}

R₂₀ for $f (x) = sin x$

on $[0, π]$

R₁₀₀ for $ln x$

on $[1, e]$

R_{100} = \frac{e - 1}{100} \sum_{i = 1}^{100} ln (1 + (e - 1) \frac{i}{100})

In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. Is the area under the curve between the left and right endpoint sums?

[T] L₁₀₀ and R₁₀₀ for $y = x^{2} - 3 x + 1$

on the interval $[−1, 1]$

[T] L₁₀₀ and R₁₀₀ for $y = x^{2}$

on the interval $[0, 1]$

A graph of the given function on the interval [0, 1]. It is set up for a left endpoint approximation and is an underestimate because the function is increasing. Ten rectangles are shown for visual clarity, but this behavior persists for more rectangles.

R_{100} = 0.33835, L_{100} = 0.32835 .

The plot shows that the left Riemann sum is an underestimate because the function is increasing. Similarly, the right Riemann sum is an overestimate. The area lies between the left and right Riemann sums. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles.

[T] L₅₀ and R₅₀ for $y = \frac{x + 1}{x^{2} - 1}$

on the interval $[2, 4]$

[T] L₁₀₀ and R₁₀₀ for $y = x^{3}$

on the interval $[−1, 1]$

A graph of the given function over [-1,1] set up for a left endpoint approximation. It is an underestimate since the function is increasing. Ten rectangles are shown for visual clarity, but this behavior persists for more rectangles.

L_{100} = −0.02, R_{100} = 0.02 .

The left endpoint sum is an underestimate because the function is increasing. Similarly, a right endpoint approximation is an overestimate. The area lies between the left and right endpoint estimates.

[T] L₅₀ and R₅₀ for $y = tan (x)$

on the interval $[0, \frac{π}{4}]$

[T] L₁₀₀ and R₁₀₀ for $y = e^{2 x}$

on the interval $[−1, 1]$

A graph of the given function over the interval -1 to 1 set up for a left endpoint approximation. It is an underestimate since the function is increasing. Ten rectangles are shown for isual clarity, but this behavior persists for more rectangles.

L_{100} = 3.555, R_{100} = 3.670 .

The plot shows that the left Riemann sum is an underestimate because the function is increasing. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles.

Let t_j denote the time that it took Tejay van Garteren to ride the jth stage of the Tour de France in 2014. If there were a total of 21 stages, interpret $\sum_{j = 1}^{21} t_{j} .$

Let $r_{j}$

denote the total rainfall in Portland on the jth day of the year in 2009. Interpret $\sum_{j = 1}^{31} r_{j} .$

The sum represents the cumulative rainfall in January 2009.

Let $d_{j}$

denote the hours of daylight and $δ_{j}$

denote the increase in the hours of daylight from day $j - 1$

to day j in Fargo, North Dakota, on the jth day of the year. Interpret $d_{1} + \sum_{j = 2}^{365} δ_{j} .$

To help get in shape, Joe gets a new pair of running shoes. If Joe runs 1 mi each day in week 1 and adds $\frac{1}{10}$

mi to his daily routine each week, what is the total mileage on Joe’s shoes after 25 weeks?

The total mileage is $7 \times \sum_{i = 1}^{25} (1 + \frac{(i - 1)}{10}) = 7 \times 25 + \frac{7}{10} \times 12 \times 25 = 385 mi .$

The following table gives approximate values of the average annual atmospheric rate of increase in carbon dioxide (CO₂) each decade since 1960, in parts per million (ppm). Estimate the total increase in atmospheric CO₂ between 1964 and 2013.

Average Annual Atmospheric CO₂ Increase, 1964–2013Source: http://www.esrl.noaa.gov/gmd/ccgg/trends/.
Decade	Ppm/y
1964–1973	1.07
1974–1983	1.34
1984–1993	1.40
1994–2003	1.87
2004–2013	2.07

The following table gives the approximate increase in sea level in inches over 20 years starting in the given year. Estimate the net change in mean sea level from 1870 to 2010.

Approximate 20-Year Sea Level Increases, 1870–1990Source: http://link.springer.com/article/10.1007%2Fs10712-011-9119-1
Starting Year	20-Year Change
1870	0.3
1890	1.5
1910	0.2
1930	2.8
1950	0.7
1970	1.1
1990	1.5

Add the numbers to get 8.1-in. net increase.

The following table gives the approximate increase in dollars in the average price of a gallon of gas per decade since 1950. If the average price of a gallon of gas in 2010 was $2.60, what was the average price of a gallon of gas in 1950?

Approximate 10-Year Gas Price Increases, 1950–2000Source: http://epb.lbl.gov/homepages/Rick\_Diamond/docs/lbnl55011-trends.pdf.
Starting Year	10-Year Change
1950	0.03
1960	0.05
1970	0.86
1980	−0.03
1990	0.29
2000	1.12

The following table gives the percent growth of the U.S. population beginning in July of the year indicated. If the U.S. population was 281,421,906 in July 2000, estimate the U.S. population in July 2010.

Annual Percentage Growth of U.S. Population, 2000–2009Source: http://www.census.gov/popest/data.
Year	% Change/Year
2000	1.12
2001	0.99
2002	0.93
2003	0.86
2004	0.93
2005	0.93
2006	0.97
2007	0.96
2008	0.95
2009	0.88

(Hint: To obtain the population in July 2001, multiply the population in July 2000 by 1.0112 to get 284,573,831.)

309,389,957

In the following exercises, estimate the areas under the curves by computing the left Riemann sums, L₈.

![A graph of a function that increases linearly with a slope of 1 from (0,1) to (3,4). It curves from (3,4) to (5,4), changing direction from increasing to decreasing at (4,5). Finally, it decreases linearly with a slope of 1 from (5,4) to (8,1).](/calculus-book/resources/CNX_Calc_Figure_05_01_201.jpg)

![The graph of a smooth curve going through the points (0,3), (1,2), (2,1), (3,2), (4,3), (5,4), (6,5), (7,4), and (8,3).](/calculus-book/resources/CNX_Calc_Figure_05_01_202.jpg)

L_{8} = 3 + 2 + 1 + 2 + 3 + 4 + 5 + 4 = 24

![The graph of a smooth curve going through the points (0,0), (1,1), (2,2), (3,1), (4,3), (5,2), (6,4), (7,5), and (8,7).](/calculus-book/resources/CNX_Calc_Figure_05_01_203.jpg)

![The graph of a smooth curve going through the points (0, 3), (1, 5), (2, 7), (3, 6), (4, 8), (5, 6), (6, 5), (7, 4), and (8, 6).](/calculus-book/resources/CNX_Calc_Figure_05_01_204.jpg)

L_{8} = 3 + 5 + 7 + 6 + 8 + 6 + 5 + 4 = 44

[T] Use a computer algebra system to compute the Riemann sum, $L_{N},$

for $N = 10, 30, 50$

for $f (x) = \sqrt{1 - x^{2}}$

on $[−1, 1] .$

[T] Use a computer algebra system to compute the Riemann sum, L_N, for $N = 10, 30, 50$

for $f (x) = \frac{1}{\sqrt{1 + x^{2}}}$

on $[−1, 1] .$

L_{10} \approx 1.7604, L_{30} \approx 1.7625, L_{50} \approx 1.76265

[T] Use a computer algebra system to compute the Riemann sum, L_N, for $N = 10, 30, 50$

for $f (x) = {sin}^{2} x$

on $[0, 2 π] .$

Compare these estimates with π.

In the following exercises, use a calculator or a computer program to evaluate the endpoint sums R_N and L_N for $N = 1,10,100 .$

How do these estimates compare with the exact answers, which you can find via geometry?

[T] $y = cos (π x)$

on the interval $[0, 1]$

R_{1} = −1, L_{1} = 1, R_{10} = −0.1, L_{10} = 0.1, L_{100} = 0.01,

and $R_{100} = −0.1 .$

By symmetry of the graph, the exact area is zero.

[T] $y = 3 x + 2$

on the interval $[3, 5]$

In the following exercises, use a calculator or a computer program to evaluate the endpoint sums R_N and L_N for $N = 1,10,100 .$

[T] $y = x^{4} - 5 x^{2} + 4$

on the interval $[−2, 2],$

which has an exact area of $\frac{32}{15}$

R_{1} = 0, L_{1} = 0, R_{10} = 2.4499, L_{10} = 2.4499, R_{100} = 2.1365, L_{100} = 2.1365

[T] $y = ln x$

on the interval $[1, 2],$

which has an exact area of $2 ln (2) - 1$

Explain why, if $f (a) \geq 0$

and f is increasing on $[a, b],$

that the left endpoint estimate is a lower bound for the area below the graph of f on $[a, b] .$

If $[c, d]$

is a subinterval of $[a, b]$

under one of the left-endpoint sum rectangles, then the area of the rectangle contributing to the left-endpoint estimate is $f (c) (d - c) .$

But, $f (c) \leq f (x)$

for $c \leq x \leq d,$

so the area under the graph of f between c and d is $f (c) (d - c)$

plus the area below the graph of f but above the horizontal line segment at height $f (c),$

which is positive. As this is true for each left-endpoint sum interval, it follows that the left Riemann sum is less than or equal to the area below the graph of f on $[a, b] .$

Explain why, if $f (b) \geq 0$

and f is decreasing on $[a, b],$

that the left endpoint estimate is an upper bound for the area below the graph of f on $[a, b] .$

Show that, in general, $R_{N} - L_{N} = (b - a) \times \frac{f (b) - f (a)}{N} .$

L_{N} = \frac{b - a}{N} \sum_{i = 1}^{N} f (a + (b - a) \frac{i - 1}{N}) = \frac{b - a}{N} \sum_{i = 0}^{N - 1} f (a + (b - a) \frac{i}{N})

and $R_{N} = \frac{b - a}{N} \sum_{i = 1}^{N} f (a + (b - a) \frac{i}{N}) .$

The left sum has a term corresponding to $i = 0$

and the right sum has a term corresponding to $i = N .$

In $R_{N} - L_{N},$

any term corresponding to $i = 1, 2,…, N - 1$

occurs once with a plus sign and once with a minus sign, so each such term cancels and one is left with $R_{N} - L_{N} = \frac{b - a}{N} (f (a + (b - a)) \frac{N}{N}) - (f (a) + (b - a) \frac{0}{N}) = \frac{b - a}{N} (f (b) - f (a)) .$

Explain why, if f is increasing on $[a, b],$

the error between either L_N or R_N and the area A below the graph of f is at most $(b - a) \frac{f (b) - f (a)}{N} .$

For each of the three graphs:

Obtain a lower bound $L (A)$
for the area enclosed by the curve by adding the areas of the squares enclosed completely by the curve.
Obtain an upper bound $U (A)$
for the area by adding to
$L (A)$
the areas
$B (A)$
of the squares enclosed partially by the curve.

Graph 1: a. $L (A) = 0, B (A) = 20;$

b. $U (A) = 20 .$

Graph 2: a. $L (A) = 9;$

b. $B (A) = 11, U (A) = 20 .$

Graph 3: a. $L (A) = 11.0;$

b. $B (A) = 4.5, U (A) = 15.5 .$

In the previous exercise, explain why $L (A)$

gets no smaller while $U (A)$

gets no larger as the squares are subdivided into four boxes of equal area.

A unit circle is made up of n wedges equivalent to the inner wedge in the figure. The base of the inner triangle is 1 unit and its height is $sin (\frac{π}{n}) .$

The base of the outer triangle is $B = cos (\frac{π}{n}) + sin (\frac{π}{n}) tan (\frac{π}{n})$

and the height is $H = B sin (\frac{2 π}{n}) .$

Use this information to argue that the area of a unit circle is equal to π.

Let A be the area of the unit circle. The circle encloses n congruent triangles each of area $\frac{sin (\frac{2 π}{n})}{2},$

so $\frac{n}{2} sin (\frac{2 π}{n}) \leq A .$

Similarly, the circle is contained inside n congruent triangles each of area $\frac{B H}{2} = \frac{1}{2} (cos (\frac{π}{n}) + sin (\frac{π}{n}) tan (\frac{π}{n})) sin (\frac{2 π}{n}),$

so $A \leq \frac{n}{2} sin (\frac{2 π}{n}) (cos (\frac{π}{n})) + sin (\frac{π}{n}) tan (\frac{π}{n}) .$

As $n \to \infty, \frac{n}{2} sin (\frac{2 π}{n}) = \frac{π sin (\frac{2 π}{n})}{(\frac{2 π}{n})} \to π,$

so we conclude $π \leq A .$

Also, as $n \to \infty, cos (\frac{π}{n}) + sin (\frac{π}{n}) tan (\frac{π}{n}) \to 1,$

so we also have $A \leq π .$

By the squeeze theorem for limits, we conclude that $A = π .$

Approximating Areas

Sigma (Summation) Notation

Proof