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1.1 Approximating Areas

Learning Objectives

  • Use sigma (summation) notation to calculate sums and powers of integers.
  • Use the sum of rectangular areas to approximate the area under a curve.
  • Use Riemann sums to approximate area.

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\left(x\right), and the x-axis on a closed interval \left[a,b\right]. 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\left(x\right) 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 \Sigma , 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

\begin{array}\ 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20.\end{array}

We could probably skip writing a couple of terms and write

1+2+3+4+\ldots+19+20,

which is better, but still cumbersome. With sigma notation, we write this sum as \ds\ds\sum\limits_{i=1}^{20}i, which is much more compact.

Sigma notation is presented in the form \ds\ds\sum\limits_{i=m}^{n}{a}_{i}, where {a}_{i} describes the terms to be added, i is the index of summation, and m\le n are limits (or bounds) of summation. Each term is evaluated, then we sum all the values, beginning with the value corresponding to i=m and ending with the one corresponding to i=n. For example, an expression like \ds\ds\sum\limits_{i=2}^{7}{s}_{i} is interpreted as {s}_{2}+{s}_{3}+{s}_{4}+{s}_{5}+{s}_{6}+{s}_{7}. 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.

Let’s try a couple of examples of using sigma notation.

Using Sigma Notation

  1. Write in sigma notation and evaluate the sum of terms {3}^{i} for i=1,2,3,4,5.
  2. Write the sum in sigma notation:

    \ds1+\ds\frac{1}{4}+\ds\frac{1}{9}+\ds\frac{1}{16}+\ds\frac{1}{25}.

Solution

  1. We have \ds\ds\sum_{i=1}^{5}{3}^{i}=3+{3}^{2}+{3}^{3}+{3}^{4}+{3}^{5}=363.
  2. The denominator of each term is a perfect square. Using sigma notation, this sum can be written as \ds\ds\sum\limits_{i=1}^{5}\ds\frac{1}{{i}^{2}}.

Write in sigma notation and evaluate the sum of terms 2^i for i=3,4,5,6.

Answer

\ds\sum\limits_{i=3}^{6}{2}^{i}=120

The properties associated with the summation process are given below. Note that because the sums arising in the definition of area below a curve and, later, the definite integral typically start from i=1, we focus on this kind of sums, but all the properties still hold for an arbitrary lower index of summation m.

Properties of Sigma Notation

Let {a}_{1},{a}_{2},\ldots,\phantom{\rule{0.2em}{0ex}}{a}_{n} and {b}_{1},{b}_{2},\ldots,\phantom{\rule{0.2em}{0ex}}{b}_{n} represent two sequences of terms and let c be a constant. The following properties hold for all positive integers n and for integers k, with 1\le k< n.

  1. \ds\ds\sum\limits_{i=1}^{n}c=nc

  2. \ds\ds\sum\limits_{i=1}^{n}c{a}_{i}=c\ds\ds\sum\limits_{i=1}^{n}{a}_{i}

  3. \ds\ds\sum\limits_{i=1}^{n}\left({a}_{i}+{b}_{i}\right)=\ds\ds\sum\limits_{i=1}^{n}{a}_{i}+\ds\ds\sum\limits_{i=1}^{n}{b}_{i}

  4. \ds\ds\sum\limits_{i=1}^{n}\left({a}_{i}-{b}_{i}\right)=\ds\ds\sum\limits_{i=1}^{n}{a}_{i}-\ds\ds\sum\limits_{i=1}^{n}{b}_{i}

  5. \ds\ds\sum\limits_{i=1}^{n}{a}_{i}=\ds\ds\sum\limits_{i=1}^{k}{a}_{i}+\ds\ds\sum\limits_{i=k+1}^{n}{a}_{i}

 

Proof

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

\begin{array}{cc}\ds 2.\ \ds\sum\limits_{i=1}^{n}c{a}_{i}\hfill &\ds =c{a}_{1}+c{a}_{2}+c{a}_{3}+\ldots+c{a}_{n}\hfill \\[5mm] &\ds =c\left({a}_{1}+{a}_{2}+{a}_{3}+\ldots+{a}_{n}\right)\hfill \\[5mm] &\ds =c\ds\ds\sum\limits_{i=1}^{n}{a}_{i}.\hfill \end{array}
\begin{array}{cc}\ds3.\ \ds\sum\limits_{i=1}^{n}\left({a}_{i}+{b}_{i}\right)\hfill &\ds =\left({a}_{1}+{b}_{1}\right)+\left({a}_{2}+{b}_{2}\right)+\left({a}_{3}+{b}_{3}\right)+\ldots+\left({a}_{n}+{b}_{n}\right)\hfill \\[5mm] &\ds =\left({a}_{1}+{a}_{2}+{a}_{3}+\ldots+{a}_{n}\right)+\left({b}_{1}+{b}_{2}+{b}_{3}+\ldots+{b}_{n}\right)\hfill \\[5mm] &\ds =\ds\ds\sum\limits_{i=1}^{n}{a}_{i}+\ds\ds\sum\limits_{i=1}^{n}{b}_{i}.\hfill \end{array}

The following formulas for sums of powers of integers simplify the summation process further, and we use them in the next set of examples.

Sums of Powers of Integers

  1. The sum of the first n integers is given by

\ds\ds\sum\limits_{i=1}^{n}i=1+2+\ldots+n=\ds\frac{n\left(n+1\right)}{2}.

  • The sum of the squares of the first n integers is given by

    \ds\ds\sum\limits_{i=1}^{n}{i}^{2}={1}^{2}+{2}^{2}+\ldots+{n}^{2}=\ds\frac{n\left(n+1\right)\left(2n+1\right)}{6}.

  • The sum of the cubes of the first n integers is given by

    \ds\ds\sum\limits_{i=1}^{n}{i}^{3}={1}^{3}+{2}^{3}+\ldots+{n}^{3}=\ds\frac{{n}^{2}{\left(n+1\right)}^{2}}{4}=\left(\ds\frac{n(n+1)}{2}\right)^2.

 

Evaluation Using Sigma Notation

Write the following sums using sigma notation and then evaluate them.

  1. The sum of the terms {\left(i-3\right)}^{2} for i=1,2,\ldots,\phantom{\rule{0.2em}{0ex}}200.

  2. The sum of the terms \left({i}^{3}-{i}^{2}\right) for i=1,2,3,4,5,6.

Solution

  1. We expand {\left(i-3\right)}^{2}, and then use properties of sigma notation along with the summation formulas to obtain

    \begin{array}{cc}\ds\ds\sum\limits_{i=1}^{200}{\left(i-3\right)}^{2}\hfill &\ds =\ds\ds\sum\limits_{i=1}^{200}\left({i}^{2}-6i+9\right)\hfill \\[5mm] &\ds =\ds\ds\sum\limits_{i=1}^{200}{i}^{2}-\ds\ds\sum\limits_{i=1}^{200}6i+\ds\ds\sum\limits_{i=1}^{200}9 \quad \text{(properties 3 and 4)}\hfill \\[5mm] &\ds =\ds\ds\sum\limits_{i=1}^{200}{i}^{2}-6\ds\ds\sum\limits_{i=1}^{200}i+\ds\ds\sum\limits_{i=1}^{200}9\quad \text{(property 2)} \hfill\\[5mm] &\ds =\ds\frac{200\left(200+1\right)\left(400+1\right)}{6}-6\left[\ds\frac{200\left(200+1\right)}{2}\right]+9\left(200\right)\hfill \\[5mm] &\ds =2,686,700-120,600+1800\hfill \\[5mm] &\ds =2,567,900\hfill \end{array}

  2. We use sigma notation property 4 and the formulas for the sum of squared terms and the sum of cubed terms to obtain

    \begin{array}{cc}\ds\ds\sum\limits_{i=1}^{6}\left({i}^{3}-{i}^{2}\right)\hfill &\ds =\ds\ds\sum\limits_{i=1}^{6}{i}^{3}-\ds\ds\sum\limits_{i=1}^{6}{i}^{2}\hfill \\[5mm] &\ds =\ds\frac{{6}^{2}{\left(6+1\right)}^{2}}{4}-\ds\frac{6\left(6+1\right)\left(2\left(6\right)+1\right)}{6}\hfill \\[5mm] &\ds =\ds\frac{1764}{4}-\ds\frac{546}{6}\hfill \\[5mm] &\ds =350\hfill \end{array}

Find the sum of the values of (4+3i) for i=1,2,\ldots,\phantom{\rule{0.2em}{0ex}}100.

Answer

15,550

Hint

Use the properties of sigma notation to solve the problem.

Finding the Sum of the Function Values

Find the sum of the values of f\left(x\right)={x}^{3} over the integers 1,2,3,\ldots,\phantom{\rule{0.2em}{0ex}}10.

Solution

Using the formula, we have

\begin{array}{cc}\ds\ds\sum\limits_{i=1}^{10}{i}^{3}\hfill &\ds =\ds\frac{{\left(10\right)}^{2}{\left(10+1\right)}^{2}}{4}\hfill \\[5mm] &\ds =\ds\frac{100\left(121\right)}{4}\hfill \\[5mm] &\ds =3025.\hfill \end{array}

Let f(x)=2x+1. Evaluate the sum \ds\sum\limits_{k=1}^{20}f\left(k\right).

Answer

440

Hint

Use the rules of sums and formulas for the sum of integers.

Approximating Area

Now that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let f\left(x\right) be a continuous, nonnegative function defined on the closed interval \left[a,b\right]. We want to approximate the area A of the region under the curve y=f\left(x\right), above the x-axis, and between the lines x=a and x=b, as shown on the figure below.

A graph in quadrant one of an area bounded by a generic curve f(x) at the top, the x-axis at the bottom, the line x = a to the left, and the line x = b to the right. About midway through, the concavity switches from concave down to concave up, and the function starts to increases shortly before the line x = b.
Figure 1. An area (shaded region) bounded by the curve f\left(x\right) at top, the x-axis at bottom, the line x=a at left, and the line x=b at right.

To approximate the area under the curve, we use a geometric approach. We divide the region into many small shapes, approximate each of them with a rectangle that has a known area formula, and then sum the areas of rectangles to obtain a reasonable estimate of the area of the region. We begin by dividing the interval \left[a,b\right] into subintervals. This step is used not only in areas, but also in definite integrals and their applications, so we formally define the relevant terminology.

Definition

Consider an interval [a,b]. A set of points P=\left\{{x}_{i}\right\}_{i=1}^n with a={x}_{0}<{x}_{1}<{x}_{2}<\ldots<{x}_{n}=b, which divides the interval \left[a,b\right] into subintervals \left[{x}_{0},{x}_{1}\right],\left[{x}_{1},{x}_{2}\right],\ldots,\phantom{\rule{0.2em}{0ex}}\left[{x}_{n-1},{x}_{n}\right] is called a partition of \left[a,b\right]. If all the subintervals have the same width, the set of points forms a regular partition of the interval \left[a,b\right]. For the regular partition, the width of each subinterval is denoted by \Delta x, so that \Delta x=\ds\frac{b-a}{n}, and then {x}_{i}={x}_{0}+i\Delta x for i=1,2,3,\ldots,\phantom{\rule{0.2em}{0ex}}n.

We use the regular partition as the basis of a method for estimating the area under the curve. The method has two variations: the left-endpoint approximation and the right-endpoint approximation, and we examine both of them below.

Left-Endpoint Approximation

On each subinterval \left[{x}_{i-1},{x}_{i}\right] (i=1,2,3,\ldots,\phantom{\rule{0.2em}{0ex}}n), construct a rectangle with a width of \Delta x and a height of f\left({x}_{i-1}\right), the function value at the left endpoint of the subinterval, which ensures that the left upper corner of the rectangle belongs to the curve y=f(x) (see Figure 2 below). This rectangle approximates the region below the graph of f over the subinterval \left[{x}_{i-1},{x}_{i}\right] and its area is f\left({x}_{i-1}\right)\Delta x. Adding the areas of all these rectangles, we get an approximate value for A. We use L_n to denote the left-endpoint approximation of A using n subintervals.

A\approx {L}_{n} =f\left({x}_{0}\right)\Delta x+f\left({x}_{1}\right)\Delta x+\ldots+f\left({x}_{n-1}\right)\Delta x=\ds\sum\limits_{i=1}^{n}f\left({x}_{i-1}\right)\Delta x

A diagram showing the left-endpoint approximation of area under a curve. Under a parabola with vertex on the y axis and above the x axis, rectangles are drawn between a=x0 on the origin and b = xn. The rectangles have endpoints at a=x0, x1, x2…x(n-1), and b = xn, spaced equally. The height of each rectangle is determined by the value of the given function at the left endpoint of the rectangle.
Figure 2. In the left-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the left endpoint of each subinterval.

The right-endpoint approximation is very similar to the left-endpoint approximation but, as the name suggests, the heights of the rectangles are determined by the function values at the right endpoint of each subinterval.

Right-Endpoint Approximation

Construct a rectangle on each subinterval \left[{x}_{i-1},{x}_{i}\right] (i=1,2,3,\ldots,\phantom{\rule{0.2em}{0ex}}n) with the height of f\left({x}_{i}\right), the function value at the right endpoint of the subinterval, which ensures that the right upper corner of the rectangle belongs to the curve y=f(x) (see Figure 3 below). Then, the area of each rectangle is f\left({x}_{i}\right)\Delta x and the approximation for A is given by

A\approx {R}_{n}=f\left({x}_{1}\right)\Delta x+f\left({x}_{2}\right)\Delta x+\ldots+f\left({x}_{n}\right)\Delta x=\ds\sum\limits_{i=1}^{n}f\left({x}_{i}\right)\Delta x.

The notation {R}_{n} indicates this is a right-endpoint approximation for A.

A diagram showing the right-endpoint approximation of area under a curve. Under a parabola with vertex on the y axis and above the x axis, rectangles are drawn between a=x0 on the origin and b = xn. The rectangles have endpoints at a=x0, x1, x2…x(n-1), and b = xn, spaced equally. The height of each rectangle is determined by the value of the given function at the right endpoint of the rectangle.
Figure 3 In the right-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the right endpoint of each subinterval. Note that the right-endpoint approximation differs from the left-endpoint approximation in Figure 2.

In Figure 4, the area of the region below the graph of the function f\left(x\right)=\ds\frac{{x}^{2}}{2} over the interval [0,3] is approximated using left- and right-endpoint approximations with six rectangles.

Diagrams side by side, showing the differences in approximating the area under a parabolic curve with vertex at the origin between the left endpoints method (the first diagram) and the right endpoints method (the second diagram). In the first diagram, rectangles are drawn at even intervals (delta x) under the curve with heights determined by the value of the function at the left endpoints. In the second diagram, the rectangles are drawn in the same fashion, but with heights determined by the value of the function at the right endpoints. The endpoints in both are spaced equally from the origin to (3, 0), labeled x0 to x6.
Figure 4. Approximating the area under a curve by using (a) the left endpoints and (b) the right endpoints.

In this case, \Delta x=\dfrac{3-0}6=0.5, and the subintervals are \left[0,0.5\right],\left[0.5,1\right],\left[1,1.5\right],\left[1.5,2\right],\left[2,2.5\right],\left[2.5,3\right], that is, x_0=0, x_1=0.5, x_2=1, x_3=1.5, x_4=2, x_5=2.5 and x_6=3. Using the left- and right-approximation formulas for L_n and R_n, we obtain

\begin{array}{ll} A&\approx {L}_{6}\ds =\ds\sum\limits_{i=1}^{6}f\left({x}_{i-1}\right)\Delta x\hfill\\[5mm] \ds&=f\left({x}_{0}\right)\Delta x+f\left({x}_{1}\right)\Delta x+f\left({x}_{2}\right)\Delta x+f\left({x}_{3}\right)\Delta x+f\left({x}_{4}\right)\Delta x+f\left({x}_{5}\right)\Delta x\hfill \\[5mm] &\ds =f\left(0\right)\cdot0.5+f\left(0.5\right)\cdot0.5+f\left(1\right)\cdot0.5+f\left(1.5\right)\cdot0.5+f\left(2\right)\cdot0.5+f\left(2.5\right)\cdot0.5\hfill \\[5mm] &\ds =0\cdot0.5+0.125\cdot0.5+0.5\cdot0.5+1.125\cdot0.5+2\cdot0.5+3.125\cdot0.5\hfill \\[5mm] &\ds =0+0.0625+0.25+0.5625+1+1.5625\hfill \\[5mm] &\ds =3.4375.\hfill \end{array}

\begin{array}{ll} A&\approx {R}_{6}\ds =\ds\sum\limits_{i=1}^{6}f\left({x}_{i}\right)\Delta x\hfill\\[5mm] \ds &=f\left({x}_{1}\right)\Delta x+f\left({x}_{2}\right)\Delta x+f\left({x}_{3}\right)\Delta x+f\left({x}_{4}\right)\Delta x+f\left({x}_{5}\right)\Delta x+f\left({x}_{6}\right)\Delta x\hfill \\[5mm] &\ds =f\left(0.5\right)\cdot0.5+f\left(1\right)\cdot0.5+f\left(1.5\right)\cdot0.5+f\left(2\right)\cdot0.5+f\left(2.5\right)\cdot0.5+f\left(3\right)\cdot0.5\hfill \\[5mm] &\ds =0.125\cdot0.5+0.5\cdot0.5+1.125\cdot0.5+2\cdot0.5+3.125\cdot0.5+4.5\cdot0.5\hfill \\[5mm] &\ds =0.0625+0.25+0.5625+1+1.5625+2.25\hfill \\[5mm] &\ds =5.6875.\hfill \end{array}

Approximating the Area Under a Curve

Use both left- and right-endpoint approximations to approximate the area under the graph of f\left(x\right)={x}^{2} over the interval \left[0,2\right] using n=4.

Solution

First, divide the interval \left[0,2\right] into n equal subintervals. Using n=4,\Delta x=\ds\frac{\left(2-0\right)}{4}=0.5. This is the width of each rectangle. The intervals \left[0,0.5\right],\left[0.5,1\right],\left[1,1.5\right],\left[1.5,2\right] are shown in Figure 5. Using the left-endpoint approximation, the heights are f\left(0\right)=0,f\left(0.5\right)=0.25,f\left(1\right)=1,f\left(1.5\right)=2.25. Then,

\begin{array}{cc}{L}_{4}\hfill &\ds =f\left({x}_{0}\right)\Delta x+f\left({x}_{1}\right)\Delta x+f\left({x}_{2}\right)\Delta x+f\left({x}_{3}\right)\Delta x\hfill \\[5mm] &\ds =0\cdot0.5+0.25\cdot0.5+1\cdot0.5+2.25\cdot0.5\hfill \\[5mm] &\ds =1.75.\hfill \end{array}

A graph of the left-endpoint approximation of the area under the curve f(x) = x^2 from 0 to 2 with endpoints spaced .5 units apart. The heights of the rectangle are determined by the values of the function at their left endpoints.
Figure 5. The graph shows the left-endpoint approximation of the area under f\left(x\right)={x}^{2} from 0 to 2.

The right-endpoint approximation is shown in the Figure 6. The intervals are the same, \Delta x=0.5, but now we use the right endpoints to calculate the heights of the rectangles. We have

\begin{array}{cc}{R}_{4}\hfill &\ds =f\left({x}_{1}\right)\Delta x+f\left({x}_{2}\right)\Delta x+f\left({x}_{3}\right)\Delta x+f\left({x}_{4}\right)\Delta x\hfill \\[5mm] &\ds =0.25\left(0.5\right)+1\left(0.5\right)+2.25\left(0.5\right)+4\left(0.5\right)\hfill \\[5mm] &\ds =3.75.\hfill \end{array}

A graph of the right-endpoint approximation method of the area under the curve f(x) = x^2 from 0 to 2 with endpoints spaced .5 units apart. The heights of the rectangles are determined by the values of the function at the right endpoints.
Figure 6. The graph shows the right-endpoint approximation of the area under f\left(x\right)={x}^{2} from 0 to 2.

According to our calculations, the left-endpoint approximation is 1.75 and the right-endpoint approximation is 3.75.

Sketch left- and right-endpoint approximations for f\left(x\right)=\ds\frac{1}{x} on \left[1,2\right] using n=4. Approximate the area using both methods.

Answer

The left-endpoint approximation is 0.7595. The right-endpoint approximation is 0.6345. See the figure below.

Two graphs side by side showing the left-endpoint approximation ad right-endpoint approximation of the area under the curve f(x) = 1/x from 1 to 2 with endpoints spaced evenly at .25 units. The heights of the left-endpoint approximation one are determined by the values of the function at the left endpoints, and the height of the right-endpoint approximation one are determined by the values of the function at the right endpoints.

Looking at the graphs above, we can see that when we use a small number of intervals, neither the left- 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 demonstrate the improved approximation obtained through smaller intervals with an example. Consider the region under the curve f\left(x\right)={\left(x-1\right)}^{3}+4 over the interval \left[0,2\right]. We first explore the left-endpoint approximations with 4 rectangles, then 8 rectangles, and finally 32 rectangles. Then, we explore the right-endpoint approximations, using the same sets of intervals.

{L}_{4}=f\left(0\right)\cdot0.5+f\left(0.5\right)\cdot0.5+f\left(1\right)\cdot0.5+f\left(1.5\right)\cdot0.5=7.5

A graph of the left-endpoint approximation of the area under the given curve from a = x0 to b=x4. The heights of the rectangles are determined by the values of the function at the left endpoints.
Figure 7. Left-endpoint approximation with 4 rectangles.

Figure 8 shows the left-endpoint approximation with 8 rectangles. Comparing to Figure 7, we can see less white space under the curve, and since the white space is a part of the region that is not included in the approximating rectangles, we get a better approximation with 8 rectangles. The area of the rectangles is

\begin{array}{cc}{L}_{8}\hfill &\ds =f\left(0\right)\cdot0.25+f\left(0.25\right)\cdot0.25+f\left(0.5\right)\cdot0.25+f\left(0.75\right)\cdot0.25\hfill \\[5mm] &\ds \phantom{\rule{0.2em}{0ex}}+f\left(1\right)\cdot0.25+f\left(1.25\right)\cdot0.25+f\left(1.5\right)\cdot0.25+f\left(1.75\right)\cdot0.25\hfill \\[5mm] &\ds =7.75.\hfill \end{array}

A graph showing the left-endpoint approximation for the area under the given curve from a=x0 to b = x8. The heights of the rectangles are determined by the values of the function at the left endpoints.
Figure 8. Left-endpoint approximation with 8 rectangles.

The graph in Figure 9 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

\begin{array}{cc}{L}_{32}\hfill &\ds =f\left(0\right)\cdot0.0625+f\left(0.0625\right)\cdot0.0625+f\left(0.125\right)\cdot0.0625+\ldots+f\left(1.9375\right)\cdot0.0625\hfill \\[5mm] &\ds =7.9375.\hfill \end{array}

A graph of the left-endpoint approximation of the area under the given curve from a = x0 to b = x32. The heights of the rectangles are determined by the values of the function at the left endpoints.
Figure 9. Here, 32 rectangles are inscribed under the curve for a left-endpoint approximation.

We can carry out a similar process for the right-endpoint approximation method. A right-endpoint approximation of the same curve, using four rectangles (as shown in Figure 10), yields an area

\begin{array}{cc}{R}_{4}\hfill &\ds =f\left(0.5\right)\cdot0.5+f\left(1\right)\cdot0.5+f\left(1.5\right)\cdot0.5+f\left(2\right)\cdot0.5\hfill \\[5mm] &\ds =8.5.\hfill \end{array}

A graph of the right-endpoint approximation for the area under the given curve from x0 to x4. The heights of the rectangles are determined by the values of the function at the right endpoints.
Figure 10. Now we divide the area under the curve into four equal subintervals for a right-endpoint approximation.

Dividing the region over the interval \left[0,2\right] into eight rectangles results in \Delta x=\ds\frac{2-0}{8}=0.25. The graph is shown in Figure 11. The area is

\begin{array}{cc}{R}_{8}\hfill &\ds =f\left(0.25\right)\cdot0.25+f\left(0.5\right)\cdot0.25+f\left(0.75\right)\cdot0.25+f\left(1\right)\cdot0.25\hfill \\[5mm] &\ds \phantom{\rule{0.2em}{0ex}}+f\left(1.25\right)\cdot0.25+f\left(1.5\right)\cdot0.25+f\left(1.75\right)\cdot0.25+f\left(2\right)\cdot0.25\hfill \\[5mm] &\ds =8.25.\hfill \end{array}

A graph of the right-endpoint approximation for the area under the given curve from a=x0 to b=x8.The heights of the rectangles are determined by the values of the function at the right endpoints.
Figure 11. Here we use right-endpoint approximation for a region divided into eight equal subintervals.

Last, the right-endpoint approximation with n=32 is close to the actual area (see Figure 12 below). The area is approximately

\begin{array}{cc}{R}_{32}\hfill &\ds =f\left(0.0625\right)\cdot0.0625+f\left(0.125\right)\cdot0.0625+f\left(0.1875\right)\cdot0.0625+\ldots+f\left(2\right)\cdot0.0625\hfill \\[5mm] &\ds =8.0625.\hfill \end{array}

A graph of the right-endpoint approximation for the area under the given curve from a=x0 to b=x32. The heights of the rectangles are determined by the values of the function at the right endpoints.
Figure 12. The region is divided into 32 equal subintervals for a right-endpoint approximation.

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. The table below 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.

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

Forming Riemann Sums

So far, to approximate the area under a curve, we have been using rectangles with the heights determined by evaluating the function at either the left or the right endpoint of the subinterval \left[{x}_{i-1},{x}_{i}\right]. However, we could evaluate the function at any point {x}_{i}^{*} in \left[{x}_{i-1},{x}_{i}\right], and use f\left({x}_{i}^{*}\right) as the height of the approximating rectangle. This would result in an estimate A\approx \ds\sum\limits_{i=1}^{n}f\left({x}_{i}^{*}\right)\Delta x.

The arising sum is called a Riemann sum, named after the 19th-century mathematician Bernhard Riemann, who developed the idea.

Riemann Sum

Let the function f\left(x\right) be defined on a closed interval \left[a,b\right] and let P be a regular partition of \left[a,b\right] with the subinterval width \Delta x. For each 1\le i\le n, let {x}_{i}^{*} be an arbitrary point in \left[{x}_{i-1},{x}_{i}\right]. The numbers x_1^*, x_2^*,\ldots, x_{n}^* are called the sample points. Then the Riemann sum for f\left(x\right) that corresponds to the partition P and the set of sample points \{x_i^*\}_{i=1}^n is defined as

\ds\sum\limits_{i=1}^{n}f\left({x}_{i}^{*}\right)\Delta x.

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

Definition

Let f\left(x\right) be a continuous, nonnegative function on an interval \left[a,b\right], and let \ds\sum\limits_{i=1}^{n}f\left({x}_{i}^{*}\right)\Delta x be a Riemann sum for f\left(x\right). Then, the area under the curve y=f\left(x\right) over \left[a,b\right] is given by

\begin{array}{c}A=\underset{n\to \infty }{\text{lim}}\ds\sum\limits_{i=1}^{n}f\left({x}_{i}^{*}\right)\Delta x.\end{array}

Some subtleties here are worth discussing. First, note the above limit is a bit different from the limit of a function g\left(x\right) as x\to\infty, because n can only take positive integer values. However, the computational techniques we used to compute limits of functions can also be used to this kind of limits.

Second, one might wonder if the limit could be different for different choices of sample points \left\{{x}_{i}^{*}\right\}. Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if f\left(x\right) is continuous on the closed interval \left[a,b\right], then \underset{n\to \infty }{\text{lim}}\ds\sum\limits_{i=1}^{n}f\left({x}_{i}^{*}\right)\Delta x exists, and it does not depend on the choice of \left\{{x}_{i}^{*}\right\}.

We now discuss some specific choices for \left\{{x}_{i}^{*}\right\}. Although any choice for \left\{{x}_{i}^{*}\right\} gives us an estimate of the area under the curve, we don’t necessarily know whether that estimate is bigger (estimate from above, or overestimate) or smaller (estimate from below, or underestimate) than the actual value of the area. However, we can always select the sample points \left\{{x}_{i}^{*}\right\} to guarantee an estimate from above or from below.

To ensure an overestimate, it is enough to choose \left\{{x}_{i}^{*}\right\} such that f\left({x}_{i}^{*}\right)\ge f\left(x\right) for all x\in \left[{x}_{i-1},{x}_{i}\right] (i=1,2,3,\ldots,n) . In other words, for each 1\le i\le n, we take {x}_{i}^{*} to be the point where f(x) attains its global maximum on the interval \left[{x}_{i-1},{x}_{i}\right] (which exists since f is continuous). In this case, the Riemann sum \ds\sum\limits_{i=1}^{n}f\left({x}_{i}^{*}\right)\Delta x is called an upper sum. Similarly, if we want an underestimate, for each 1\le i\le n, we can choose {x}_{i}^{*} to be the point where f(x) attains its global minimum on the interval \left[{x}_{i-1},{x}_{i}\right]. The associated Riemann sum is called a lower sum. Note that if f\left(x\right) is either increasing or decreasing throughout the interval \left[a,b\right], 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.

Finding Lower and Upper Sums

Find the lower sum for f\left(x\right)=10-{x}^{2} over \left[1,2\right] corresponding to n=4 subintervals.

Solution

We have that \Delta x=\ds=\frac{2-1}4=\frac{1}{4}, and the intervals are \left[1,1.25\right],\left[1.25,1.5\right],\left[1.5,1.75\right],\left[1.75,2\right]. Because the function is decreasing over the interval \left[1,2\right], the lower sum is obtained by using the right endpoints of the subintervals as the sample points, that is, it is equal to the right-endpoint approximation.

The graph of f(x) = 10 − x^2 from 0 to 2. It is set up for a right-end approximation of the area bounded by the curve and the x axis on [1, 2], labeled a=x0 to x4. It shows a lower sum.
Figure 13. Right-endpoint approximation with n=4.
\begin{array}{ll}R_4&\ds=\ds\sum\limits_{k=1}^{4}\left(10-{x_i}^{2}\right)\cdot0.25\\[5mm] &=0.25\left[10-{\left(1.25\right)}^{2}+10-{\left(1.5\right)}^{2}+10-{\left(1.75\right)}^{2}+10-{\left(2\right)}^{2}\right]\\[5mm] &\ds =0.25\left[8.4375+7.75+6.9375+6\right]\hfill \\[5mm] &\ds =7.28.\hfill \end{array}

Hence, the lower sum is 7.28.

Find the upper sum for f\left(x\right)=10-{x}^{2} over \left[1,2\right] corresponding to n=4 subintervals and sketch the corresponding approximation.

Answer

The upper sum is 8.0313.

A graph of the function f(x) = 10 − x^2 from 0 to 2. It is set up for a right endpoint approximation over the area [1,2], which is labeled a=x0 to x4. It is an upper sum.

Hint

f\left(x\right) is decreasing on \left[1,2\right], so the maximum function values occur at the left endpoints of the subintervals.

Finding Lower and Upper Sums

Find the lower sum for f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}(x) over \left[a,b\right]=\left[0,\ds\frac{\pi }{2}\right] corresponding to n=6 subintervals.

Solution

We have that \Delta x=\dfrac{\pi/2-0}{6}=\dfrac{\pi}{12}, and the intervals are \left[0,\ds\frac{\pi }{12}\right],\left[\ds\frac{\pi }{12},\ds\frac{\pi }{6}\right],\left[\ds\frac{\pi }{6},\ds\frac{\pi }{4}\right],\left[\ds\frac{\pi }{4},\ds\frac{\pi }{3}\right],\left[\ds\frac{\pi }{3},\ds\frac{5\pi }{12}\right], and \left[\ds\frac{5\pi }{12},\ds\frac{\pi }{2}\right]. Since f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}(x) is increasing on the interval \left[0,\ds\frac{\pi }{2}\right], for each subinterval, the function attains its minimum value at the left endpoint, and so the lower sum coincides with the left-endpoint approximation.

A graph of the function y = sin(x) from 0 to pi. It is set up for a left endpoint approximation from 0 to pi/2 and n=6. It is a lower sum.
Figure 14. The left-endpoint approximation with 6 subintervals.
\begin{array}{ll}L_4 &\ds =\text{sin}\left(0\right)\cdot\ds\frac{\pi }{12}+\text{sin}\left(\ds\frac{\pi }{12}\right)\cdot\ds\frac{\pi }{12}+\text{sin}\left(\ds\frac{\pi }{6}\right)\cdot\ds\frac{\pi }{12}\\[5mm]&+\text{sin}\left(\ds\frac{\pi }{4}\right)\cdot\ds\frac{\pi }{12}+\text{sin}\left(\ds\frac{\pi }{3}\right)\cdot\ds\frac{\pi }{12}+\text{sin}\left(\ds\frac{5\pi }{12}\right)\cdot\ds\frac{\pi }{12}\hfill \\[5mm] &\ds =\frac{\pi }{12}\left[0+\text{sin}\left(\ds\frac{\pi }{12}\right)+\dfrac12+\dfrac{\sqrt2}2+\dfrac{\sqrt3}2+\text{sin}\left(\ds\frac{5\pi }{12}\right)\right]\hfill \end{array}

So we obtain that the lower sum is equal to the above expression. Note that, using the trigonometric formula \sin(\alpha)+\sin(\beta)=2\sin\left(\dfrac{\alpha+\beta}{2}\right)\cos\left(\dfrac{\alpha-\beta}{2}\right), we can obtain that \text{sin}\left(\ds\frac{\pi }{12}\right)+\text{sin}\left(\ds\frac{5\pi }{12}\right)=2\sin\left(\dfrac{\pi}{4}\right)\cos\left(\dfrac{-\pi}6\right)=\dfrac{\sqrt 6}2, and hence the answer can be simplified as

\dfrac{\pi(1+\sqrt2+\sqrt3+\sqrt6)}{24}.

Find the upper sum for f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}(x) over \left[a,b\right]=\left[0,\ds\frac{\pi }{2}\right] corresponding to n=6 subintervals.

Answer

\dfrac{\pi(3+\sqrt2+\sqrt3+\sqrt6)}{24}

Hint

Compare the expression for the upper sum to the one for the lower sum, and use the answer to the previous example.

Key Concepts

  • Sigma (summation) notation of the form \ds\sum\limits_{i=m}^{n}{a}_{i} is useful for expressing long sums of values in compact form.
  • For a continuous function defined over an interval \left[a,b\right], the process of dividing the interval into n equal parts, building a rectangle on each of them with the height determined by the value of the function at the left (respectively, right) endpoint of the subinterval, calculating the areas of these rectangles, and then summing the areas yields a left-endpoint (respectively, right-endpoint) approximation of the area below the graph of f.
  • The width of each rectangle is \Delta x=\ds\frac{b-a}{n}.
  • Riemann sums are expressions of the form \ds\sum\limits_{i=1}^{n}f\left({x}_{i}^{*}\right)\Delta x, and they can be used to estimate the area under the curve y=f\left(x\right). Left- and right-endpoint approximations are particular cases of Riemann sums.
  • Riemann sums allow for much flexibility in choosing the set of points \left\{{x}_{i}^{*}\right\} at which the function is evaluated, often with an eye to obtaining a lower sum or an upper sum.

Key Equations

  • Properties of Sigma Notation
    \ds\sum\limits_{i=1}^{n}c=nc

    \ds\sum\limits_{i=1}^{n}c{a}_{i}=c\ds\sum\limits_{i=1}^{n}{a}_{i}\ds\sum\limits_{i=1}^{n}\left({a}_{i}+{b}_{i}\right)=\ds\sum\limits_{i=1}^{n}{a}_{i}+\ds\sum\limits_{i=1}^{n}{b}_{i}

    \ds\sum\limits_{i=1}^{n}\left({a}_{i}-{b}_{i}\right)=\ds\sum\limits_{i=1}^{n}{a}_{i}-\ds\sum\limits_{i=1}^{n}{b}_{i}

    \ds\sum\limits_{i=1}^{n}{a}_{i}=\ds\sum\limits_{i=1}^{k}{a}_{i}+\ds\sum\limits_{i=k+1}^{n}{a}_{i}

  • Sums and Powers of Integers
    \ds\sum\limits_{i=1}^{n}i=1+2+\ldots+n=\ds\frac{n\left(n+1\right)}{2}

    \ds\sum\limits_{i=1}^{n}{i}^{2}={1}^{2}+{2}^{2}+\ldots+{n}^{2}=\ds\frac{n\left(n+1\right)\left(2n+1\right)}{6}

    \ds\sum\limits_{i=0}^{n}{i}^{3}={1}^{3}+{2}^{3}+\ldots+{n}^{3}=\ds\frac{{n}^{2}{\left(n+1\right)}^{2}}{4}

  • Left-Endpoint Approximation
    A\approx {L}_{n}=f\left({x}_{0}\right)\Delta x+f\left({x}_{1}\right)\Delta x+\ldots+f\left({x}_{n-1}\right)\Delta x=\ds\sum\limits_{i=1}^{n}f\left({x}_{i-1}\right)\Delta x
  • Right-Endpoint Approximation
    A\approx {R}_{n}=f\left({x}_{1}\right)\Delta x+f\left({x}_{2}\right)\Delta x+\ldots+f\left({x}_{n}\right)\Delta x=\ds\sum\limits_{i=1}^{n}f\left({x}_{i}\right)\Delta x

Exercises

1. State whether the given sums are equal or unequal.

  1. \ds\sum\limits_{i=1}^{10}i and \ds\sum\limits_{k=1}^{10}k

  2. \ds\sum\limits_{i=1}^{10}i and \ds\sum\limits_{i=6}^{15}\left(i-5\right)

  3. \ds\sum\limits_{i=1}^{10}i\left(i-1\right) and \ds\sum\limits_{j=0}^{9}\left(j+1\right)j

  4. \ds\sum\limits_{i=1}^{10}i\left(i-1\right) and \ds\sum\limits_{k=1}^{10}\left({k}^{2}-k\right)

Answer

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 formulas for sums of powers of integers and the 5-th property of sigma notation to compute the sums.

2. \ds\ds\sum\limits_{i=5}^{10}i

3. \ds\sum\limits_{i=5}^{10}{i}^{2}

Answer

385-30=355

Suppose that \ds\sum\limits_{i=1}^{100}{a}_{i}=15 and \ds\sum\limits_{i=1}^{100}{b}_{i}=-12. In the following exercises, compute the sums.

4. \ds\sum\limits_{i=1}^{100}\left({a}_{i}+{b}_{i}\right)

5. \ds\sum\limits_{i=1}^{100}\left({a}_{i}-{b}_{i}\right)

Answer

15-\left(-12\right)=27

6. \ds\sum\limits_{i=1}^{100}\left(3{a}_{i}-4{b}_{i}\right)

7. \ds\sum\limits_{i=1}^{100}\left(5{a}_{i}+4{b}_{i}\right)

Answer

5\left(15\right)+4\left(-12\right)=27

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

8. \ds\sum\limits_{k=1}^{10}\left({k}^{3}+1\right)

9. \ds\sum\limits_{j=1}^{50}\left({j}^{2}-2j\right)

Answer

\ds\sum\limits_{j=1}^{50}{j}^{2}-2\ds\sum\limits_{j=1}^{50}j=\ds\frac{\left(50\right)\left(51\right)\left(101\right)}{6}-\ds\frac{2\left(50\right)\left(51\right)}{2}=40,\text{​}375

10. \ds\sum\limits_{j=1}^{8}\left(3-j\right)(3+j)

11. \ds\sum\limits_{k=1}^{25}4k(k-25)

Answer

\ds\sum\limits_{k=1}^{25}\left[4k^2-100k\right]=\ds\frac{4\left(25\right)\left(26\right)\left(51\right)}{6}-50\left(25\right)\left(26\right)=-10,400

Let {L}_{n} and {R}_{n} denote, respectively, the left- and the right-endpoint approximations of the area under y=f(x) over the interval [a,b]. In the following exercises, compute the indicated left- and right-endpoint approximations of the areas corresponding to the given functions and intervals.

12. L_4 for f\left(x\right)=\ds\frac{1}{x-1} over \left[2,3\right]

13. R_4 for f\left(x\right)=\text{cos}\left(\pi x\right) over \left[0,1\right]

Answer

{R}_{4}=-0.25

14. L_6 for f\left(x\right)=\ds\frac{1}{x+2} over \left[-1,2\right]

15. R_6 for f\left(x\right)=\ds\frac{1}{x\left(x-1\right)} over \left[2,5\right]

Answer

{R}_{6}=0.372

16. R_3 for \ds f(x)=\sin(2x) over \left[-\pi,0\right]

17. L_4 for \ds f(x)=\frac{1}{{x}^{2}+1} over \left[-2,2\right]

Answer

{L}_{4}=2.20

18. L_3 for f(x)=3-x^2 over \left[1,4\right]

19. L_8 for f(x)={x}^{2}-2x+1 over \left[0,2\right]

Answer

{L}_{8}=0.6875

20. Compute the left and right Riemann sums — L_4 and R_4, respectively — for f\left(x\right)=2-|x| over \left[-2,2\right]. Compute their average value and compare it with the area under the graph of f.

21. Compute the left and right Riemann sums — L_6 and R_6, respectively — for f\left(x\right)=3-|3-x| over \left[0,6\right]. Compute their average value and compare it with the area under the graph of f.

Answer

{L}_{6}=9={R}_{6}. The graph of f is a triangle with area 9.

22. Compute the left and right Riemann sums — L_4 and R_4, respectively — for f\left(x\right)=\sqrt{4-{x}^{2}} over \left[-2,2\right] and compare their values.

23. Compute the left and right Riemann sums — L_6 and R_6, respectively — for f\left(x\right)=\sqrt{9-{\left(x-3\right)}^{2}} over \left[0,6\right] and compare their values.

Answer

{L}_{6}=13.12899={R}_{6}. They are equal.

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

24. L_{30} for f\left(x\right)={x}^{2} over \left[1,2\right]

25. L_{10} for f\left(x\right)=\sqrt{4-{x}^{2}} over \left[-2,2\right]

Answer

{L}_{10}=\ds\frac{4}{10}\ds\sum\limits_{i=1}^{10}\sqrt{4-\left(-2+4\ds\frac{\left(i-1\right)}{10}\right)^2}

26. R_{20} for f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}(x) over \left[0,\pi \right]

27. R_{100} for f(x)=\text{ln}\phantom{\rule{0.1em}{0ex}}(x) over \left[1,e\right]

Answer

{R}_{100}=\ds\frac{e-1}{100}\ds\sum\limits_{i=1}^{100}\text{ln}\left(1+\left(e-1\right)\ds\frac{i}{100}\right)

28. Let t_j denote the time that it took Tejay van Garteren to ride the j-th stage of the Tour de France in 2014. If there were a total of 21 stages, interpret \ds\sum\limits_{j=1}^{21}{t}_{j}.

29. Let {r}_{j} denote the total rainfall in Portland on the j-th day of the year in 2009. Interpret \ds\sum\limits_{j=1}^{31}{r}_{j}.

Answer

The sum represents the cumulative rainfall in January 2009.

30. Let {d}_{j} denote the hours of daylight and {\delta }_{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}+\ds\sum\limits_{j=2}^{365}{\delta }_{j}.

31. 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 \ds\frac{1}{10} mi to his daily routine each week, what is the total mileage on Joe’s shoes after 25 weeks?

Answer

7\cdot\ds\sum\limits_{i=1}^{25}\left(1+\ds\frac{\left(i-1\right)}{10}\right)=7\cdot 25+\ds\frac{7}{10}\cdot12\cdot25=385\ \text{mi}.

In the following exercises, estimate the areas under the curves by computing the left Riemann sums, L_8.

32.
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).

33.
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).

Answer

{L}_{8}=3+2+1+2+3+4+5+4=24

34.
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).

35. Explain why, if f\left(a\right)\ge 0 and f is increasing on \left[a,b\right], then the left-endpoint approximation is a lower bound for the area below the graph of f on \left[a,b\right].

Solution

If \left[x_j,x_{j+1}\right] is a subinterval of \left[a,b\right], then since f is increasing, we have that f\left(x_j\right)\le f\left(x\right) for x_j\le x\le x_{j+1}. This means that the approximating rectangle corresponding to the subinterval \left[x_j,x_{j+1}\right] is completely inside the region below the graph of f over this subinterval, and hence the area of the rectangle is no bigger than the area of this region. As this is true for each approximating rectangle, when we sum up their areas, we obtain that the left Riemann sum is less than or equal to the area below the graph of f over \left[a,b\right].

36. Explain why, if f\left(b\right)\ge 0 and f is decreasing on \left[a,b\right], then the left-endpoint approximation is an upper bound for the area below the graph of f over \left[a,b\right].

37. Show that, in general, {R}_{N}-{L}_{N}=\left(b-a\right)\cdot\ds\frac{f\left(b\right)-f\left(a\right)}{N}.

Solution

{L}_{N}=\ds\frac{b-a}{N}\ds\sum\limits_{i=1}^{N}f\left(a+\left(b-a\right)\ds\frac{i-1}{N}\right)=\ds\frac{b-a}{N}\ds\sum\limits_{i=0}^{N-1}f\left(a+\left(b-a\right)\ds\frac{i}{N}\right) and {R}_{N}=\ds\frac{b-a}{N}\ds\sum\limits_{i=1}^{N}f\left(a+\left(b-a\right)\ds\frac{i}{N}\right). 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,\ldots,\phantom{\rule{0.2em}{0ex}}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}=\ds\frac{b-a}{N}\left(f\left(a+\left(b-a\right)\right)\ds\frac{N}{N}\right)-\left(f\left(a\right)+\left(b-a\right)\ds\frac{0}{N}\right)=\ds\frac{b-a}{N}\left(f\left(b\right)-f\left(a\right)\right).

38. Explain why, if f is increasing on \left[a,b\right], the error between either L_n or R_n and the area A below the graph of f is at most \left(b-a\right)\ds\frac{f\left(b\right)-f\left(a\right)}{N}.

39. For each of the three graphs:

  1. Obtain a lower bound L\left(A\right) for the area enclosed by the curve by adding the areas of the squares enclosed completely by the curve.
  2. Obtain an upper bound U\left(A\right) for the area by adding to L\left(A\right) the areas B\left(A\right) of the squares enclosed partially by the curve.
     Three graphs, stacked vertically, drawn on graph paper. Each shows the same image. However, the axes become progressively more exact in units. The first is marked in units, from negative 3 units to positive 3 units on each axis. The second has the half-units marked, and the third has the quarter units marked. As such, the graph paper boxes become smaller and smaller. The image is symmetrical across each axis and is a curved cross shape. It meets the axes at (0,3), (3,0), (0,-3), and (-3,0) and has corners roughly at (.7,.7), (.7,-.7), (-.7,-7.), and (-.7,.7). In graph 1, no square unit boxes are completely contained inside the shape. Twenty boxes are enclosed partially by the shape. In graph 2, nine boxes are completely contained inside the shape, and eleven boxes are enclosed partially by the shape. In graph 3, 11 boxes are completely contained inside the shape, and 4.5 are enclosed partially by the shape.

Answer

Graph 1: a. L\left(A\right)=0,B\left(A\right)=20; b. U\left(A\right)=20. Graph 2: a. L\left(A\right)=9; b. B\left(A\right)=11,U\left(A\right)=20. Graph 3: a. L\left(A\right)=11.0; b. B\left(A\right)=4.5,U\left(A\right)=15.5.

40. In the previous exercise, explain why L\left(A\right) gets no smaller while U\left(A\right) gets no larger as the squares are subdivided into four boxes of equal area.

41. 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 \text{sin}\left(\ds\frac{\pi }{n}\right). The base of the outer triangle is B=\text{cos}\left(\ds\frac{\pi }{n}\right)+\text{sin}\left(\ds\frac{\pi }{n}\right)\text{tan}\left(\ds\frac{\pi }{n}\right) and the height is H=B\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\ds\frac{2\pi }{n}\right). Use this information to argue that the area of a unit circle is equal to \pi.

A wedge of a circle cut at an acute angle theta = 2pi / n. Several extra lines are drawn. The first is a line A connecting the ends of the two radii, creating a triangle. The second is another line B parallel to the A, connecting the radii a few units in from each endpoint. A concentric curve C connects the endpoints of B and is tangent to A near its midpoint. The area between this curve C and the edge of the circle is shaded in pink, and the rest of the wedge is purple. A final concentric curve is drawn very close to angle theta.

Solution

Let A be the area of the unit circle. The circle encloses n congruent triangles each of area \ds\frac{\text{sin}\left(\ds\frac{2\pi }{n}\right)}{2}, so \ds\frac{n}{2}\,\text{sin}\left(\ds\frac{2\pi }{n}\right)\le A. Similarly, the circle is contained inside n congruent triangles each of area \ds\frac{BH}{2}=\ds\frac{1}{2}\left(\text{cos}\left(\ds\frac{\pi }{n}\right)+\text{sin}\left(\ds\frac{\pi }{n}\right)\text{tan}\left(\ds\frac{\pi }{n}\right)\right)\text{sin}\left(\ds\frac{2\pi }{n}\right), and so A\le \ds\frac{n}{2}\,\text{sin}\left(\ds\frac{2\pi }{n}\right)\left(\text{cos}\left(\ds\frac{\pi }{n}\right)+\text{sin}\left(\ds\frac{\pi }{n}\right)\text{tan}\left(\ds\frac{\pi }{n}\right)\right). As n\to \infty, we have that \dfrac{2\pi}n\to0, implying that \ds\frac{n}{2}\,\text{sin}\left(\ds\frac{2\pi }{n}\right)=\ds\frac{\pi \phantom{\rule{0.1em}{0ex}}\text{sin}\left(\ds\frac{2\pi }{n}\right)}{\left(\ds\frac{2\pi }{n}\right)}\to \pi , and hence \pi \le A. We also have that \text{cos}\left(\ds\frac{\pi }{n}\right)+\text{sin}\left(\ds\frac{\pi }{n}\right)\text{tan}\left(\ds\frac{\pi }{n}\right)\to \cos(0)+\sin(0)\tan(0)=1, as n\to\infty, which yields A\le \pi . Therefore, A=\pi .

Glossary

sigma notation
(also, summation notation) the Greek letter sigma (\ds\sigma) indicates addition of the values; the values of the index above and below the sigma indicate where to begin the summation and where to end it
partition
a set of points that divides an interval into subintervals
regular partition
a partition in which the subintervals all have the same width
left-endpoint approximation
an approximation of the area under a curve computed by using the left endpoint of each subinterval to calculate the height of the approximating rectangle
right-endpoint approximation
an approximation of the area under a curve using the right endpoint of each subinterval to calculate the height of the approximating rectangle
riemann sum
the sum of the form \ds\sum\limits_{i=1}^{n}f\left({x}_{i}^{*}\right)\Delta x, where each x_i^* is a sample point in the subinterval [x_{i-1},x_i]
lower sum
a Riemann sum obtained by using the minimum value of f\left(x\right) on each subinterval
upper sum
a Riemann sum obtained by using the maximum value of f\left(x\right) on each subinterval

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Calculus: Volume 2 (Second University of Manitoba Edition) Copyright © 2021 by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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