Riemann Sums from a Table

The integral of a rate

Suppose

r(t)

is the rate at which some quantity changes, in units per unit of time. Over a short time interval of length

\Delta t

the rate hardly changes, so the quantity changes by about

r(t)\,\Delta t

. A Riemann sum adds these small changes across

[a, b]

, and in the limit the sum becomes the integral.

Where the rate is negative the quantity decreases, and the integral subtracts that decrease. So the integral is an amount gained minus an amount lost, not a total of everything that happened.

Equal widths

When the table lists the rate at equally spaced times, every subinterval has the same width

\Delta t

. The sums of the earlier lessons then apply as they stand.

Example 1

Rain falls at a rate of

r(t)

millimeters per hour, where

t

is measured in hours, and the table gives selected values. Approximate the integral of

r

from

0

8

with a left sum, a right sum, and a trapezoidal sum, using the four subintervals the table gives. Say what the result means.

$t$ (hours)	$0$	$2$	$4$	$6$	$8$
$r(t)$ (millimeters per hour)	$4$	$7$	$9$	$6$	$3$

The five times give four subintervals, each of width

\Delta t = 2

. A left sum reads each rate at the start of its subinterval, so it uses every value but the last. A right sum uses every value but the first:

L = 2(4 + 7 + 9 + 6) = 52, \qquad R = 2(7 + 9 + 6 + 3) = 50.

In the trapezoidal sum each interior value counts twice and each end value once:

T = \frac{2}{2}\bigl[4 + 2(7) + 2(9) + 2(6) + 3\bigr] = 51,

which is the average of the two endpoint sums,

\frac{52 + 50}{2} = 51

. Rain can't fall at a negative rate, so here the net change is the whole amount: about

51

millimeters of rain fell in the

8

hours.

Unequal widths

Measurements are often taken at uneven times. Then there is no common width to factor out, and each term of the sum carries its own width. The rest of this lesson uses one table of that kind.

A cyclist rides along a straight road. The velocity

v(t)

, in meters per minute, is positive when the cyclist rides east and negative when the cyclist rides west, where

t

is measured in minutes. The function

v

is continuous, and selected values are shown.

$t$ (minutes)	$0$	$3$	$5$	$9$	$15$	$18$
$v(t)$ (meters per minute)	$260$	$200$	$150$	$80$	$-40$	$-100$

Example 2

Approximate the integral of

v

from

0

18

with a left sum and a right sum, using the five subintervals the table gives. Explain what the integral means.

The widths are

3

2

4

6

, and

3

. Each term is a width in minutes times a velocity in meters per minute, so each estimates a change in position, in meters. Write every product before adding:

\begin{aligned} L &= 3(260) + 2(200) + 4(150) + 6(80) + 3(-40) \\[4pt] &= 780 + 400 + 600 + 480 - 120 = 2140, \\[10pt] R &= 3(200) + 2(150) + 4(80) + 6(-40) + 3(-100) \\[4pt] &= 600 + 300 + 320 - 240 - 300 = 680. \end{aligned}

The two usual mistakes are multiplying each velocity by its time instead of by its width, and spreading the widths evenly as

\frac{18}{5}

when they aren't equal.

The integral of

v

from

0

18

is the net change in the cyclist's position, in meters, over those

18

minutes. It tells where the ride ends relative to the start, with east positive and west negative, and it isn't the distance ridden. The velocity changes sign between

t = 9

and

t = 15

, and the integral counts the distance ridden west as negative.

Example 3

Approximate the same integral with a trapezoidal sum using the five subintervals.

The Trapezoidal Rule's pattern of coefficients needs equal widths, so compute each trapezoid on its own: its width times the average of its two end values.

\begin{aligned} T &= 3\cdot\frac{260 + 200}{2} + 2\cdot\frac{200 + 150}{2} + 4\cdot\frac{150 + 80}{2} + 6\cdot\frac{80 + (-40)}{2} + 3\cdot\frac{-40 + (-100)}{2} \\[8pt] &= 690 + 350 + 460 + 120 - 210 = 1410. \end{aligned}

Each trapezoid is the average of its left and right rectangles. So even with unequal widths, the trapezoidal sum is the average of the two sums in Example 2:

\frac{2140 + 680}{2} = 1410

A midpoint sum from a table

A midpoint sum needs the rate at the middle of each subinterval. A table supplies it only when the midpoint is one of the listed times, so the subintervals are chosen to fit the table. They are usually fewer and wider than the table's own.

Too big or too small

The table shows

v

only at six times, and says nothing about what it does in between. So a claim about an overestimate or an underestimate needs more information about

v

than the table gives.