CC Riemann Sums

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Section 4.2 Riemann Sums

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Motivating Questions

How can we use a Riemann sum to estimate the area between a given curve and the horizontal axis over a particular interval?
What are the differences among left, right, middle, and random Riemann sums?
How can we write Riemann sums in an abbreviated form?

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Supplemental Videos.

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The main topics of this section are also presented in the following videos:

Riemann Sums Part 1
¹⁴
unl.yuja.com/V/Video?v=7114334&node=34303273&a=155756359&autoplay=1
Riemann Sums Part 2
¹⁵
unl.yuja.com/V/Video?v=7114333&node=34303231&a=201197937&autoplay=1
Riemann Sums Examples
¹⁶
unl.yuja.com/V/Video?v=7114331&node=34303262&a=172730896&autoplay=1

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In Section 4.1, we learned that if an object moves with positive velocity

v,

the area between

y = v (t)

and the

t

-axis over a given time interval tells us the distance traveled by the object over that time interval. If

v (t)

is sometimes negative and we view the area of any region below the

t

-axis as having an associated negative sign, then the sum of these signed areas tells us the moving object’s change in position over a given time interval.

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Consider the velocity function given in Figure 4.27. If the areas of shaded regions are

A_{1},

A_{2},

and

A_{3}

as labeled, then the total distance

D

traveled by the moving object on

[a, b]

D = A_{1} + A_{2} + A_{3},

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while the total change in the object’s position on

[a, b]

s (b) - s (a) = A_{1} - A_{2} + A_{3} .

Figure 4.27. A velocity function that is sometimes negative.

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Because the motion is in the negative direction on the interval where

v (t) < 0,

we subtract

A_{2}

to determine the object’s total change in position.

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Of course, finding

D

and

s (b) - s (a)

for the graph in Figure 4.27 presumes that we can actually find the areas

A_{1},

A_{2},

and

A_{3} .

So far, we have worked with velocity functions that were either constant or linear, so that the area bounded by the velocity function and the horizontal axis is a combination of rectangles and triangles, and we can find the area exactly. However, if the curve bounds a region that is not a familiar geometric shape, then we cannot find its area exactly. Indeed, this is one of our biggest goals in Chapter 4: to learn how to find the exact area bounded between a curve and the horizontal axis for as many different types of functions as possible.

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In Example 4.8, we approximated the area under a nonlinear velocity function using rectangles. In Example 4.28, we consider three different options for the heights of the rectangles we will use.

Example 4.28.

A person walking along a straight path has her velocity in miles per hour at time

t

given by the function

v (t) = 0.25 t^{3} - 1.5 t^{2} + 3 t + 0.25,

for times in the interval

0 \leq t \leq 2 .

The graph of this function is also given in each of the three diagrams in Figure 4.29.

Figure 4.29. Three approaches to estimating the area under $y = v (t)$ on the interval $[0, 2] .$

Note that in each diagram, we use four rectangles to estimate the area under

y = v (t)

on the interval

[0, 2],

but the method by which the four rectangles’ respective heights are decided varies among the three individual graphs.

How are the heights of rectangles in the left-most diagram being chosen? Explain, and hence determine the value of

$S = A_{1} + A_{2} + A_{3} + A_{4}$

by evaluating the function $y = v (t)$ at appropriately chosen values and observing the width of each rectangle. Note, for example, that

$A_{3} = v (1) \cdot \frac{1}{2} = 2 \cdot \frac{1}{2} = 1 .$
Explain how the heights of rectangles are being chosen in the middle diagram and find the value of

$T = B_{1} + B_{2} + B_{3} + B_{4} .$
Likewise, determine the pattern of how heights of rectangles are chosen in the right-most diagram and determine

$U = C_{1} + C_{2} + C_{3} + C_{4} .$
Of the estimates $S,$ $T,$ and $U,$ which do you think is the best approximation of $D,$ the total distance the person traveled on $[0, 2] ?$ Why?

Hint.

On which sides of the rectangles do the red dots appear?

Answer.

$S = 2.9375$
$B = 3.9375$
$U$ appears to be the most accurate estimate.

Solution.

The heights of the rectangles are chosen by evaluating the function $v (t) = 0.25 t^{3} - 1.5 t^{2} + 3 t + 0.25$ at the left-hand endpoints. Specifically,

$\begin{aligned} h_{1} & = v (0) = 0.25 (0)^{3} - 1.5 (0)^{2} + 3 (0) + 0.25 = 0.25 \\ h_{2} & = v (0.5) = 0.25 (0)^{3} - 1.5 (0)^{2} + 3 (0) + 0.25 = 1.40625 \\ h_{3} & = v (1) = 0.25 (0)^{3} - 1.5 (0)^{2} + 3 (0) + 0.25 = 2 \\ h_{4} & = v (1.5) = 0.25 (0)^{3} - 1.5 (0)^{2} + 3 (0) + 0.25 = 2.21875 \end{aligned}$

Recall that the width of each rectangle is $0.5 .$ To determine $S,$ we compute

$A_{1} + A_{2} + A_{3} + A_{4} .$

Then,

$\begin{aligned} A_{1} & = v (0) \cdot 0.5 = 0.25 \cdot 0.5 = 0.125 \\ A_{2} & = v (0.5) \cdot 0.5 = 1.40625 \cdot 0.5 = 0.703125 \\ A_{3} & = v (1) \cdot 0.5 = 2 \cdot 0.5 = 1 \\ A_{4} & = v (1.5) \cdot 0.5 = 2.21875 \cdot 0.5 = 1.109375 \end{aligned}$

Putting it all together,

$\begin{aligned} S & = A_{1} + A_{2} + A_{3} + A_{4} & = 0.125 + 0.703125 + 1 + 1.109375 & = 2.9375 \end{aligned}$

Therefore,

$S = 2.9375$
The heights of the rectangles are chosen by evaluating the function $v (t) = 0.25 t^{3} - 1.5 t^{2} + 3 t + 0.25$ at the right-hand endpoints. Specifically,

$\begin{aligned} h_{1} & = v (0.5) = 0.25 (0.5)^{3} - 1.5 (0.5)^{2} + 3 (0.5) + 0.25 = 1.40625 \\ h_{2} & = v (1) = 0.25 (1)^{3} - 1.5 (1)^{2} + 3 (1) + 0.25 = 2 \\ h_{3} & = v (1.5) = 0.25 (1.5)^{3} - 1.5 (1.5)^{2} + 3 (1.5) + 0.25 = 2.21875 \\ h_{4} & = v (2) = 0.25 (2)^{3} - 1.5 (2)^{2} + 3 (2) + 0.25 = 2.25 \end{aligned}$

The width of each rectangle is $0.5 .$ To determine $T,$ we compute

$B_{1} + B_{2} + B_{3} + B_{4} .$

Then,

$\begin{aligned} B_{1} & = v (0.5) \cdot 0.5 = 1.40625 \cdot 0.5 = 0.703125 \\ B_{2} & = v (1) \cdot 0.5 = 2 \cdot 0.5 = 1 \\ B_{3} & = v (1.5) \cdot 0.5 = 2.21875 \cdot 0.5 = 1.109375 \\ B_{4} & = v (2) \cdot 0.5 = 2.25 \cdot 0.5 = 1.125 \end{aligned}$

Putting it all together,

$\begin{aligned} T & = B_{1} + B_{2} + B_{3} + B_{4} & = 0.703125 + 1 + 1.109375 + 1.125 & = 3.9375 \end{aligned}$

Therefore,

$T = 3.9375$
The heights of the rectangles are chosen by evaluating the function $v (t) = 0.25 t^{3} - 1.5 t^{2} + 3 t + 0.25$ at the points in-between the right-hand and left-hand endpoints i.e. the midpoint. Specifically,

$\begin{aligned} h_{1} & = v (0.25) = 0.25 (0.25)^{3} - 1.5 (0.25)^{2} + 3 (0.25) + 0.25 = 0.91016 \\ h_{2} & = v (0.75) = 0.25 (0.75)^{3} - 1.5 (0.75)^{2} + 3 (0.75) + 0.25 = 1.7617 \\ h_{3} & = v (1.25) = 0.25 (1.25)^{3} - 1.5 (1.25)^{2} + 3 (1.25) + 0.25 = 2.1445 \\ h_{4} & = v (1.75) = 0.25 (1.75)^{3} - 1.5 (1.75)^{2} + 3 (1.75) + 0.25 = 2.2461 \end{aligned}$

The width of each rectangle is $0.5 .$ To determine $U,$ we compute

$C_{1} + C_{2} + C_{3} + C_{4} .$

Then,

$\begin{aligned} C_{1} & = v (0.25) \cdot 0.5 = 0.91016 \cdot 0.5 = 0.45508 \\ C_{2} & = v (0.75) \cdot 0.5 = 1.7617 \cdot 0.5 = 0.88085 \\ C_{3} & = v (1.25) \cdot 0.5 = 2.1445 \cdot 0.5 = 1.07225 \\ C_{4} & = v (1.75) \cdot 0.5 = 2.2461 \cdot 0.5 = 1.12305 \end{aligned}$

Putting it all together,

$\begin{aligned} U & = C_{1} + C_{2} + C_{3} + C_{4} & = 0.45508 + 0.88085 + 1.07225 + 1.12305 & = 3.53123 \end{aligned}$

Therefore,

$U = 3.53123$
Visual inspection of the graphs makes it look like that $U$ is the best approximation. The estimate $S$ is clearly an under-estimate based on the graph; this can be further justified by the computation done in part (a). The estimate $T$ is clearly an over-estimate based on the graph; this can be further justified by the computation done in part (b). Hence, the estimate $U$ is a fairly accurate approximation.

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Subsection 4.2.1 Sigma Notation

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We have used sums of areas of rectangles to approximate the area under a curve. Intuitively, we expect that using a larger number of thinner rectangles will provide a better estimate for the area. Consequently, we anticipate dealing with sums of a large number of terms. To do so, we introduce sigma notation, named for the Greek letter

Σ,

which is the capital letter

S

in the Greek alphabet.

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For example, say we are interested in the sum

1 + 2 + 3 + \dots + 100,

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i.e. the sum of the first 100 natural numbers. In sigma notation, we write

\sum_{k = 1}^{100} k = 1 + 2 + 3 + \dots + 100 .

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We read the symbol

\sum_{k = 1}^{100} k

as “the sum from

k

equals 1 to 100 of

k .

” The variable

k

is called the index of summation, and any letter can be used for this variable. The pattern in the terms of the sum is denoted by a function of the index; for example,

\sum_{k = 1}^{10} (k^{2} + 2 k) = (1^{2} + 2 \cdot 1) + (2^{2} + 2 \cdot 2) + (3^{2} + 2 \cdot 3) + \dots + (10^{2} + 2 \cdot 10),

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and more generally,

\sum_{k = 1}^{n} f (k) = f (1) + f (2) + \dots + f (n) .

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Sigma notation allows us to vary easily the function being used to describe the terms in the sum, and to adjust the number of terms in the sum simply by changing the value of

n .

We test our understanding of this new notation in the following example.

Example 4.30.

For each sum written in sigma notation, write the sum long-hand and evaluate the sum to find its value. For each sum written in expanded form, write the sum in sigma notation.

$\sum_{k = 1}^{5} (k^{2} + 2)$
$\sum_{i = 3}^{6} (2 i - 1)$
$3 + 7 + 11 + 15 + \dots + 27$
$4 + 8 + 16 + 32 + \dots + 256$
$\sum_{i = 1}^{6} \frac{1}{2^{i}}$

Hint.

Observe that when $k = 1,$ $k^{2} + 2 = 1^{2} + 2 = 3 .$ This is the first term in the sum.
Note that this sum starts at $i = 3 .$
Since the terms in the sum increase by 4, try a function $f (k)$ that somehow involves $4 k .$
What pattern do you observe in the terms of the sum?
Write every term in the sum as a fraction with denominator $2^{6} = 64 .$

Answer.

$64$
$32$
$3 + 7 + 11 + 15 + \dots + 27 = \sum_{k = 1}^{7} 4 k - 1 .$
$4 + 8 + 16 + 32 + \dots + 256 = \sum {i = 2}^{8} 2^{i} .$
$\sum_{i = 1}^{6} \frac{1}{2^{i}} = \frac{63}{64} .$

Solution.

$\begin{aligned} \sum_{k = 1}^{5} (k^{2} + 2) = & (1^{2} + 2) + (2^{2} + 2) + (3^{2} + 2) + (4^{2} + 2) + (5^{2} + 2) \\ = & 3 + 6 + 11 + 17 + 27 \\ = & 64 \end{aligned}$
$\begin{aligned} \sum_{i = 3}^{6} (2 i - 1) = & (2 \cdot 3 - 1) + (2 \cdot 4 - 1) + (2 \cdot 5 - 1) + (2 \cdot 6 - 1) \\ = & 5 + 7 + 9 + 11 \\ = & 32 \end{aligned}$
Observe that each term in the sum

$3 + 7 + 11 + 15 + \dots + 27$

differs from the previous term by 4. If we view $4$ as $4 = 4 \cdot 1 - 1$ and $7$ as $7 = 4 \cdot 2 - 1,$ we see that the pattern may be represented through the function $f (k) = 4 k - 1,$ so that

$3 + 7 + 11 + 15 + \dots + 27 = \sum_{k = 1}^{7} 4 k - 1 .$

We note that $k = 7$ is the end value of the index since $4 \cdot 7 = 28 .$
The sum $4 + 8 + 16 + 32 + \dots + 256$ is a sum of powers of $2,$ which we can express in sigma notation as

$4 + 8 + 16 + 32 + \dots + 256 = \sum {i = 2}^{8} 2^{i} .$
$\begin{aligned} \sum_{i = 1}^{6} \frac{1}{2^{i}} = & \frac{1}{2} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{6}} \\ = & \frac{32}{64} + \frac{16}{64} + \frac{8}{64} + \frac{4}{64} + \frac{2}{64} + \frac{1}{64} \\ = & \frac{63}{64} . \end{aligned}$

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Subsection 4.2.2 Riemann Sums

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When a moving body has a positive velocity function

y = v (t)

on a given interval

[a, b],

the area under the curve over the interval gives the total distance the body travels on

[a, b] .

We are also interested in finding the exact area bounded by

y = f (x)

on an interval

[a, b],

regardless of the meaning or context of the function

f .

For now, we continue to focus on finding an accurate estimate of this area by using a sum of the areas of rectangles. Unless otherwise indicated, we assume that

f

is continuous and non-negative on

[a, b] .

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The first choice we make in such an approximation is the number of rectangles.

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Figure 4.31. Subdividing the interval $[a, b]$ into $n$ subintervals of equal length $Δ x .$

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If we desire

n

rectangles of equal width to subdivide the interval

[a, b],

then each rectangle must have width

Δ x = \frac{b - a}{n} .

We let

x_{0} = a,

x_{n} = b,

and define

x_{i} = a + i Δ x,

so that

x_{1} = x_{0} + Δ x,

x_{2} = x_{0} + 2 Δ x,

and so on, as pictured in Figure 4.31.

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We use each subinterval

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

as the base of a rectangle, and next choose the height of the rectangle on that subinterval. There are three standard choices: we can use the left endpoint of each subinterval, the right endpoint of each subinterval, or the midpoint of each. These are precisely the options encountered in Example 4.28 and seen in Figure 4.29. We next explore how these choices can be described in sigma notation.

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Consider an arbitrary positive function

f

[a, b]

with the interval subdivided as shown in Figure 4.31, and choose to use left endpoints. Then on each interval

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

the area of the rectangle formed is given by

A_{i + 1} = f (x_{i}) \cdot Δ x,

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as seen in Figure 4.32.

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Figure 4.32. Subdividing the interval $[a, b]$ into $n$ subintervals of equal length $Δ x$ and approximating the area under $y = f (x)$ over $[a, b]$ using left rectangles.

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If we let

LEFT (n)

denote the sum of the areas of these rectangles, we see that

\begin{aligned} LEFT (n) = & A_{1} + A_{2} + \dots + A_{i + 1} + \dots + A_{n} \\ = & f (x_{0}) \cdot Δ x + f (x_{1}) \cdot Δ x + \dots + f (x_{i}) \cdot Δ x + \dots + f (x_{n - 1}) \cdot Δ x . \end{aligned}

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In the more compact sigma notation, we have

LEFT (n) = \sum_{i = 0}^{n - 1} f (x_{i}) Δ x .

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Note that since the index of summation begins at

0

and ends at

n - 1,

there are indeed

n

terms in this sum. We call

LEFT (n)

the left Riemann sum for the function

f

on the interval

[a, b] .

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To see how the Riemann sums for right endpoints and midpoints are constructed, we consider Figure 4.33.

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Figure 4.33. Riemann sums using right endpoints and midpoints.

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For the sum with right endpoints, we see that the area of the rectangle on an arbitrary interval

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

is given by

B_{i + 1} = f (x_{i + 1}) \cdot Δ x,

and that the sum of all such areas of rectangles is given by

\begin{aligned} RIGHT (n) = & B_{1} + B_{2} + \dots + B_{i + 1} + \dots + B_{n} \\ = & f (x_{1}) \cdot Δ x + f (x_{2}) \cdot Δ x + \dots + f (x_{i + 1}) \cdot Δ x + \dots + f (x_{n}) \cdot Δ x \\ = & \sum_{i = 1}^{n} f (x_{i}) Δ x . \end{aligned}

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We call

RIGHT (n)

the right Riemann sum for the function

f

on the interval

[a, b] .

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For the sum that uses midpoints, we introduce the notation

{\overset{―}{x}}_{i + 1} = \frac{x_{i} + x_{i + 1}}{2}

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so that

{\overset{―}{x}}_{i + 1}

is the midpoint of the interval

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

For instance, for the rectangle with area

C_{1}

in Figure 4.33, we now have

C_{1} = f ({\overset{―}{x}}_{1}) \cdot Δ x .

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Hence, the sum of all the areas of rectangles that use midpoints is

\begin{aligned} MID (n) = & C_{1} + C_{2} + \dots + C_{i + 1} + \dots + C_{n} \\ = & f (\overset{―}{x_{1}}) \cdot Δ x + f (\overset{―}{x_{2}}) \cdot Δ x + \dots + f ({\overset{―}{x}}_{i + 1}) \cdot Δ x + \dots + f ({\overset{―}{x}}_{n}) \cdot Δ x \\ = & \sum_{i = 1}^{n} f ({\overset{―}{x}}_{i}) Δ x, \end{aligned}

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and we say that

MID (n)

is the middle Riemann sum for

f

[a, b] .

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Thus, we have two variables to explore: the number of rectangles and the height of each rectangle. We can explore these choices dynamically, and the applet

¹⁷

Marc Renault, Geogebra Calculus Applets.

found at http://gvsu.edu/s/a9

¹⁸

gvsu.edu/s/a9

is a particularly useful one. There we see the image shown in Figure 4.34, but with the opportunity to adjust the slider bars for the heights and the number of rectangles.

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By moving the sliders, we can see how the heights of the rectangles change as we consider left endpoints, midpoints, and right endpoints, as well as the impact that a larger number of narrower rectangles has on the approximation of the exact area bounded by the function and the horizontal axis.

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When

f (x) \geq 0

[a, b],

each of the Riemann sums

LEFT (n),

RIGHT (n),

and

MID (n)

provides an estimate of the area under the curve

y = f (x)

over the interval

[a, b] .

We also recall that in the context of a nonnegative velocity function

y = v (t),

the corresponding Riemann sums approximate the distance traveled on

[a, b]

by a moving object with velocity function

v .

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There is a more general way to think of Riemann sums, and that is to allow any choice of where the function is evaluated to determine the rectangle heights. Rather than saying we’ll always choose left endpoints, or always choose midpoints, we simply say that a point

x_{i + 1}^{*}

will be selected at random in the interval

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

(so that

x_{i} \leq x_{i + 1}^{*} \leq x_{i + 1}

). The Riemann sum is then given by

f (x_{1}^{*}) \cdot Δ x + f (x_{2}^{*}) \cdot Δ x + \dots + f (x_{i + 1}^{*}) \cdot Δ x + \dots + f (x_{n}^{*}) \cdot Δ x = \sum_{i = 1}^{n} f (x_{i}^{*}) Δ x .

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At http://gvsu.edu/s/a9

²⁰

gvsu.edu/s/a9

, the applet noted earlier and referenced in Figure 4.34, by unchecking the “relative” box at the top left, and instead checking “random,” we can easily explore the effect of using random point locations in subintervals on a Riemann sum. In computational practice, we most often use

LEFT (n),

RIGHT (n),

MID (n),

while the random Riemann sum is useful in theoretical discussions. In the following example, we investigate several different Riemann sums for a particular velocity function.

Example 4.35.

Suppose that an object moving along a straight line path has its velocity in feet per second at time

t

in seconds given by

v (t) = \frac{2}{9} (t - 3)^{2} + 2 .

Carefully sketch the region whose exact area will tell you the value of the distance the object traveled on the time interval $2 \leq t \leq 5 .$
Estimate the distance traveled on $[2, 5]$ by computing $LEFT (4),$ $RIGHT (4),$ and $MID (4) .$
Does averaging $LEFT (4)$ and $RIGHT (4)$ result in the same value as $MID (4) ?$ If not, what do you think the average of $LEFT (4)$ and $RIGHT (4)$ measures?
For this question, think about an arbitrary function $f,$ rather than the particular function $v$ given above. If $f$ is positive and increasing on $[a, b],$ will $LEFT (n)$ over-estimate or under-estimate the exact area under $f$ on $[a, b] ?$ Will $RIGHT (n)$ over- or under-estimate the exact area under $f$ on $[a, b] ?$ Explain.

Hint.

Note that $y = v (t)$ is a parabola with vertex $(3, 2) .$
Recall the formulas for $LEFT (n),$ $RIGHT (n),$ and $MID (n) .$
Think about what the average of $LEFT (1)$ and $RIGHT (1)$ measures.
Consider carefully the role of endpoints in generating $LEFT (n)$ and $RIGHT (n) .$

Answer.

$LEFT (4) = \frac{311}{48} \approx 6.47917,$ $RIGHT (4) = \frac{335}{48} \approx 6.97917,$ and $MID (4) = \frac{637}{96} \approx 6.63542 .$
$\frac{LEFT (4) + MID (4)}{2} = \frac{646}{96} \neq \frac{637}{96} = MID (4) .$
$LEFT (n)$ is an under-estimate; $RIGHT (n)$ is an over-estimates.

Solution.

The region whose exact area tells us the value of the distance the object traveled on the time interval $2 \leq t \leq 5$ is shown below.
$LEFT (4) = \frac{311}{48} \approx 6.47917,$ $RIGHT (4) = \frac{335}{48} \approx 6.97917,$ and $MID (4) = \frac{637}{96} \approx 6.63542 .$
The average of $LEFT (4)$ and $RIGHT (4)$ is

$\frac{LEFT (4) + MID (4)}{2} = \frac{311 + 335}{96} = \frac{646}{96} \neq \frac{637}{96} = MID (4) .$

This average actually measures what would result from using four trapezoids, rather than rectangles, to estimate the area on each subinterval. One reason this is so is because the area of a trapezoid is the average of the bases times the width, and the “bases” are given by the function values at the left and right endpoints.
If $f$ is positive and increasing on $[a, b],$ $LEFT (n)$ will under-estimate the exact area under $f$ on $[a, b] .$ Because $f$ is increasing, its value at the left endpoint of any subinterval will be lower than every other function value in the interval, and thus the rectangle with that height lies exclusively below the curve. In a similar way, $RIGHT (n)$ over-estimates the exact area under $f$ on $[a, b] .$

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Subsection 4.2.3 When the function is sometimes negative

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For a Riemann sum such as

LEFT (n) = \sum_{i = 0}^{n - 1} f (x_{i}) Δ x,

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we can of course compute the sum even when

f

takes on negative values. We know that when

f

is positive on

[a, b],

a Riemann sum estimates the area bounded between

f

and the horizontal axis over the interval.

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Figure 4.36. At left and center, two left Riemann sums for a function $f$ that is sometimes negative; at right, the areas bounded by $f$ on the interval $[a, d] .$

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For the function pictured in the first graph of Figure 4.36, a left Riemann sum with 12 subintervals over

[a, d]

is shown. The function is negative on the interval

b \leq x \leq c,

so at the four left endpoints that fall in

[b, c],

the terms

f (x_{i}) Δ x

are negative. This means that those four terms in the Riemann sum produce an estimate of the opposite of the area bounded by

y = f (x)

and the

x

-axis on

[b, c] .

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In the middle graph of Figure 4.36, we see that by increasing the number of rectangles the approximation of the area (or the opposite of the area) bounded by the curve appears to improve.

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In general, any Riemann sum of a continuous function

f

on an interval

[a, b]

approximates the difference between the area that lies above the horizontal axis on

[a, b]

and under

f

and the area that lies below the horizontal axis on

[a, b]

and above

f .

In the notation of Figure 4.36, we may say that

LEFT (24) \approx A_{1} - A_{2} + A_{3},

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where

LEFT (24)

is the left Riemann sum using 24 subintervals shown in the middle graph.

A_{1}

and

A_{3}

are the areas of the regions where

f

is positive, and

A_{2}

is the area where

f

is negative. We will call the quantity

A_{1} - A_{2} + A_{3}

the net signed area bounded by

f

over the interval

[a, d],

where by the phrase “signed area” we indicate that we are attaching a minus sign to the areas of regions that fall below the horizontal axis.

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Finally, we recall that if the function

f

represents the velocity of a moving object, the sum of the areas bounded by the curve tells us the total distance traveled over the relevant time interval, while the net signed area bounded by the curve computes the object’s change in position on the interval.

Example 4.37.

Suppose that an object moving along a straight line path has its velocity

v

(in feet per second) at time

t

(in seconds) given by

v (t) = \frac{1}{2} t^{2} - 3 t + \frac{7}{2} .

Compute $MID (5),$ the middle Riemann sum, for $v$ on the time interval $[1, 5] .$ Be sure to clearly identify the value of $Δ t$ as well as the locations of $t_{0},$ $t_{1},$ $\dots,$ $t_{5} .$ In addition, provide a careful sketch of the function and the corresponding rectangles that are being used in the sum.
Building on your work in (a), estimate the total change in position of the object on the interval $[1, 5] .$
Building on your work in (a) and (b), estimate the total distance traveled by the object on $[1, 5] .$
Use appropriate computing technology
²¹
For instance, consider the applet at http://gvsu.edu/s/a9 and change the function and adjust the locations of the blue points that represent the interval endpoints $a$ and $b .$
to compute $MID (10)$ and $MID (20) .$ What exact value do you think the middle sum eventually approaches as $n$ increases without bound? What does that number represent in the physical context of the overall problem?

Hint.

Note that $Δ t = \frac{5 - 1}{5} .$
Change in position is tied to the net signed area bounded by the velocity function.
Think about how total distance is different from change in position when the velocity is sometimes negative.
Besides the noted applet, computer algebra systems such as Maple and Mathematica offer this utility.

Answer.

$MID (5) = - \frac{36}{25} = - 1.44$
The change in position is approximately $- 1.44$ feet.
$D \approx 2.336 .$
$- \frac{4}{3}$ is the object’s total change in position on $[1, 5] .$

Solution.

For this Riemann sum with five subintervals, $Δ t = \frac{5 - 1}{5} = \frac{4}{5},$ so $t_{0} = 1,$ $t_{1} = 1.8,$ $t_{2} = 2.6,$ $t_{3} = 3.4,$ $t_{4} = 4.2$ and $t_{5} = 4 .$ It follows that

$\begin{aligned} MID (5) = & v (1.4) Δ t + v (2.2) Δ t + v (3.0) Δ t + v (3.8) Δ t + v (4.6) \\ = & \frac{7}{25} \cdot \frac{4}{5} - \frac{17}{25} \cdot \frac{4}{5} - 1 \cdot \frac{4}{5} - \frac{17}{25} \cdot \frac{4}{5} + \frac{7}{25} \cdot \frac{4}{5} \\ = & - \frac{36}{25} = - 1.44 \end{aligned}$

A sketch is shown below.
Since the net signed area bounded by $v$ on $[1, 5]$ represents the total change in position of the object on the interval $[1, 5],$ it follows that $MID (5)$ estimates the total change in position. Hence, the change in position is approximately $- 1.44$ feet.
To estimate the total distance traveled by the object on $[1, 5],$ we have to calculate the total area between the curve and the $t$ -axis. Thus,

$D \approx \frac{7}{25} \cdot \frac{4}{5} + \frac{17}{25} \cdot \frac{4}{5} + 1 \cdot \frac{4}{5} + \frac{17}{25} \cdot \frac{4}{5} + \frac{7}{25} \cdot \frac{4}{5} = \frac{292}{125} \approx 2.336 .$
Using appropriate technology, $MID (10) = - 1.36$ and $MID (20) = - 1.34 .$ Further calculations suggest that $MID (n) \to - \frac{4}{3} = - 1. \overset{―}{33}$ as $n \to \infty,$ and this number represents the object’s total change in position on $[1, 5] .$

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Subsection 4.2.4 Summary

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A Riemann sum is simply a sum of products of the form $f (x_{i}^{*}) Δ x$ that estimates the area between a positive function and the horizontal axis over a given interval. If the function is sometimes negative on the interval, the Riemann sum estimates the difference between the areas that lie above the horizontal axis and those that lie below the axis.
The three most common types of Riemann sums are left, right, and middle sums, but we can also work with a more general Riemann sum. The only difference among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed. For a left Riemann sum, we evaluate the function at the left endpoint of each subinterval, while for right and middle sums, we use right endpoints and midpoints, respectively.
The left, right, and middle Riemann sums are denoted $LEFT (n),$ $RIGHT (n),$ and $MID (n),$ with formulas

$\begin{aligned} LEFT (n) = f (x_{0}) Δ x + f (x_{1}) Δ x + \dots + f (x_{n - 1}) Δ x & = \sum_{i = 0}^{n - 1} f (x_{i}) Δ x, \\ RIGHT (n) = f (x_{1}) Δ x + f (x_{2}) Δ x + \dots + f (x_{n}) Δ x & = \sum_{i = 1}^{n} f (x_{i}) Δ x, \\ MID (n) = f ({\overset{―}{x}}_{1}) Δ x + f ({\overset{―}{x}}_{2}) Δ x + \dots + f ({\overset{―}{x}}_{n}) Δ x & = \sum_{i = 1}^{n} f ({\overset{―}{x}}_{i}) Δ x, \end{aligned}$

where $x_{0} = a,$ $x_{i} = a + i Δ x,$ and $x_{n} = b,$ using $Δ x = \frac{b - a}{n} .$ For the midpoint sum, ${\overset{―}{x}}_{i} = \frac{x_{i - 1} + x_{i}}{2} .$

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Exercises 4.2.5 Exercises

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1. Evaluating Riemann sums for a quadratic function.

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The rectangles in the graph below illustrate a left endpoint Riemann sum for

f (x) = \frac{- x^{2}}{5} + 2 x

on the interval

[4, 8] .

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The value of this left endpoint Riemann sum is , and this Riemann sum is

[select an answer]
an overestimate of
equal to
an underestimate of
there is ambiguity

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the area of the region enclosed by

y = f (x),

the x-axis, and the vertical lines x = 4 and x = 8.

Left endpoint Riemann sum for

y = \frac{- x^{2}}{5} + 2 x

[4, 8]

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The rectangles in the graph below illustrate a right endpoint Riemann sum for

f (x) = \frac{- x^{2}}{5} + 2 x

on the interval

[4, 8] .

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The value of this right endpoint Riemann sum is , and this Riemann sum is

[select an answer]
an overestimate of
equal to
an underestimate of
there is ambiguity

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the area of the region enclosed by

y = f (x),

the x-axis, and the vertical lines x = 4 and x = 8.

Right endpoint Riemann sum for

y = \frac{- x^{2}}{5} + 2 x

[4, 8]

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2. Estimating distance traveled with a Riemann sum from data.

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Your task is to estimate how far an object traveled during the time interval

0 \leq t \leq 8,

but you only have the following data about the velocity of the object.

time (sec)	0	1	2	3	4	5	6	7	8
velocity (feet/sec)	3	1	3	4	-1	-2	-3	-2	-1

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To get an idea of what the velocity function might look like, you pick up a black pen, plot the data points, and connect them by curves. Your sketch looks something like the black curve in the graph below.

Left endpoint approximation

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You decide to use a left endpoint Riemann sum to estimate the total displacement. So, you pick up a blue pen and draw rectangles whose height is determined by the velocity measurement at the left endpoint of each one-second interval. By using the left endpoint Riemann sum as an approximation, you are assuming that the actual velocity is approximately constant on each one-second interval (or, equivalently, that the actual acceleration is approximately zero on each one-second interval), and that the velocity and acceleration have discontinuous jumps every second. This assumption is probably incorrect because it is likely that the velocity and acceleration change continuously over time. However, you decide to use this approximation anyway since it seems like a reasonable approximation to the actual velocity given the limited amount of data.

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(A) Using the left endpoint Riemann sum, find approximately how far the object traveled. Your answers must include the correct units

²²

/webwork2_files/helpFiles/Units.html

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Total displacement =

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Total distance traveled =

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Using the same data, you also decide to estimate how far the object traveled using a right endpoint Riemann sum. So, you sketch the curve again with a black pen, and draw rectangles whose height is determined by the velocity measurement at the right endpoint of each one-second interval.

Right endpoint approximation

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(B) Using the right endpoint Riemann sum, find approximately how far the object traveled. Your answers must include the correct units

²³

/webwork2_files/helpFiles/Units.html

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Total displacement =

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Total distance traveled =

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3. Writing basic Riemann sums.

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On a sketch of

y = \ln (x),

represent the left Riemann sum with

n = 2

approximating

\int_{2}^{3} \ln (x) d x .

Write out the terms of the sum, but do not evaluate it:

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Sum = +

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On another sketch, represent the right Riemann sum with

n = 2

approximating

\int_{2}^{3} \ln (x) d x .

Write out the terms of the sum, but do not evaluate it:

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Sum = +

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Which sum is an overestimate?

the left Riemann sum
the right Riemann sum
neither sum

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Which sum is an underestimate?

the left Riemann sum
the right Riemann sum
neither sum

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4. Using the Middle Riemann sum.

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Consider the function

f (x) = 3 x + 4 .

Compute $MID (4)$ for $y = f (x)$ on the interval $[2, 5] .$ Be sure to clearly identify the value of $Δ x,$ as well as the locations of $x_{0}, x_{1}, \dots, x_{4} .$ Include a careful sketch of the function and the corresponding rectangles being used in the sum.
Use a familiar geometric formula to determine the exact value of the area of the region bounded by $y = f (x)$ and the $x$ -axis on $[2, 5] .$
Explain why the values you computed in (a) and (b) turn out to be the same. Will this be true if we use a number different than $n = 4$ and compute $MID (n) ?$ Will $LEFT (4)$ or $RIGHT (4)$ have the same value as the exact area of the region found in (b)?
Describe the collection of functions $g$ for which it will always be the case that $MID (n),$ regardless of the value of $n,$ gives the exact net signed area bounded between the function $g$ and the $x$ -axis on the interval $[a, b] .$

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5. Identifying and manipulating Riemann sum components.

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Let

S

be the sum given by

S = ((1.4)^{2} + 1) \cdot 0.4 + ((1.8)^{2} + 1) \cdot 0.4 + ((2.2)^{2} + 1) \cdot 0.4 + ((2.6)^{2} + 1) \cdot 0.4 .

Assume that $S$ is a right Riemann sum. For what function $f$ and what interval $[a, b]$ is $S$ this function’s Riemann sum? Why?
How does your answer to (a) change if $S$ is a left Riemann sum? a middle Riemann sum?
Suppose that $S$ really is a right Riemann sum. What geometric quantity does $S$ approximate?
Use sigma notation to write a new sum $R$ that is the right Riemann sum for the same function, but that uses twice as many subintervals as $S .$

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6. Evaluating Riemann sums with data.

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A car traveling along a straight road is braking and its velocity is measured at several different points in time, as given in the following table.

Table 4.38. Data for the braking car.

seconds, $t$	$0$	$0.3$	$0.6$	$0.9$	$1.2$	$1.5$	$1.8$
Velocity in ft/sec, $v (t)$	$100$	$88$	$74$	$59$	$40$	$19$	$0$

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Plot the given data on a set of axes with time on the horizontal axis and the velocity on the vertical axis.
Estimate the total distance traveled during the car the time brakes using a middle Riemann sum with 3 subintervals.
Estimate the total distance traveled on $[0, 1.8]$ by computing $LEFT (6),$ $RIGHT (6),$ and $\frac{1}{2} (LEFT (6) + RIGHT (6)) .$
Assuming that $v (t)$ is always decreasing on $[0, 1.8],$ what is the maximum possible distance the car traveled before it stopped? Why?

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7. Evaluating Riemann sums with graphs and formulas.

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The rate at which pollution escapes a scrubbing process at a manufacturing plant increases over time as filters and other technologies become less effective. For this particular example, assume that the rate of pollution (in tons per week) is given by the function

r

that is pictured in Figure 4.39.

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Use the graph to estimate the value of $MID (4)$ on the interval $[0, 4] .$
What is the meaning of $MID (4)$ in terms of the pollution discharged by the plant?
Suppose that $r (t) = 0.5 e^{0.5 t} .$ Use this formula for $r$ to compute $LEFT (5)$ on $[0, 4] .$
Determine an upper bound on the total amount of pollution that can escape the plant during the pictured four week time period that is accurate within an error of at most one ton of pollution.

Figure 4.39. The rate, $r (t),$ of pollution in tons per week.

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