Unit 6 · Topic 6.2 Beta

Estimating Area with Riemann Sums

A Riemann sum estimates accumulated change by adding width times height over subintervals; a trapezoidal sum averages the two end heights.

Practice 1: Implementing Mathematical ProcessesPractice 2: Connecting Representations

Question set for this topic

Part 1 · Hook

Why this matters

A rain gauge logs the flow into a tank only every hour or two, and some readings are three hours apart. Nobody knows how fast it flowed between readings, yet the city still needs to know how much water was collected. Rectangles and trapezoids built on the readings give an honest estimate, and the shape of the data tells you whether that estimate is too high or too low.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Water flows at a constant 30 liters per minute for 4 minutes. How much water flows?

120 liters
34 liters
7.5 liters
30 liters

Show the answer

Rate times time is the area of a rectangle under the rate graph: 30 times 4.

Correct: 120 liters:
34 liters:
7.5 liters:
30 liters:

2. If $f'(x) > 0$ on $[a, b]$ , where on $[a, b]$ is $f$ smallest?

at $x = a$
at $x = b$
in the middle
it depends on the concavity

Show the answer

$f' > 0$ means $f$ is increasing, so its smallest value on the interval is at the left end.

Correct: at $x = a$ :
at $x = b$ :
in the middle:
it depends on the concavity:

3. The graph of $f$ is concave down. Where does a chord between two points of the graph lie?

below the graph
above the graph
on the graph
it depends on the slope

Show the answer

A concave down graph bends downward, so it bulges above every chord.

Correct: below the graph:
above the graph:
on the graph:
it depends on the slope:

Part 4 · See it

See it first

Four panels show f(x) = 0.25x squared + 1 on [0, 4], increasing and concave up, with four subintervals of width 1: a left sum of 7.5 (an underestimate), a right sum of 11.5 (an overestimate), a midpoint sum of 9.25 (an underestimate) and a trapezoidal sum of 9.5 (an overestimate). The true area is 28/3, about 9.33. — Left, right, midpoint and trapezoidal sums on the same increasing, concave up function, with four subintervals of width 1. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

You only know a rate at a few timestreat it as roughly constant on each subinterval and add up width times height
The table has times 0, 1, 3, 4, 7the widths are 1, 2, 1 and 3; each product uses its own width
The function is increasing on the intervalleft ends give the smallest heights, so a left sum underestimates and a right sum overestimates
The graph is concave upchords lie above the graph, so a trapezoidal sum overestimates

Part 6 · Key ideas

Key ideas

A Riemann sum adds (width) times (height) over the subintervals: left, right or midpoint heights.
A trapezoidal sum uses width times the average of the two end heights; it equals (left + right) / 2.
From a table, each width is the difference of two times. Never assume equal widths.
Increasing function: left sum under, right sum over. Decreasing: the reverse. Concave up: trapezoid over, midpoint under.

Part 7 · Misconception

A common mistake

The wrong idea: With 4 subintervals on $[0, 7]$ , every width is $\frac{7}{4}$ .

What actually happens: Only when the subintervals are equal. A table with times 0, 1, 3, 4, 7 has widths 1, 2, 1 and 3, and each product must use its own width.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Rainwater flowing into a tank

Rainwater flows into a storage tank at a rate

R(t)

gallons per hour, where

t

is measured in hours.

R

is increasing on

0 \le t \le 7

. Selected values are in the table.

Rate R(t), in gallons per hour, at selected times t, in hours
$t$ (hours)	0	1	3	4	7
$R(t)$ (gallons per hour)	20	26	30	35	41

1. Use a left Riemann sum with the four subintervals indicated by the table to estimate the gallons of rainwater that flow into the tank from $t = 0$ to $t = 7$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

Left sum: $1(20)$ $+ 2(26)$ $+ 1(30)$ $+ 3(35)$ $= 20$ $+ 52$ $+ 30$ $+ 105$ $= 207$ gallons.

Answer: 207 gallons

2. Which expression is the right Riemann sum with the four subintervals indicated by the table?

$1(20) + 2(26) + 1(30) + 3(35)$
$1.75(26)$ $+ 1.75(30)$ $+ 1.75(35)$ $+ 1.75(41)$
$1(26) + 3(30) + 4(35) + 7(41)$
$1(26) + 2(30) + 1(35) + 3(41)$

Show the answer

Widths: $1 - 0 = 1$ , $3 - 1 = 2$ , $4 - 3 = 1$ , $7 - 4 = 3$ . Right ends: $R(1) = 26$ , $R(3) = 30$ , $R(4) = 35$ , $R(7) = 41$ . Right sum $= 26 + 60 + 35 + 123$ $= 244$ gallons.

$1(20) + 2(26) + 1(30) + 3(35)$ : These heights are the rates at the left ends, so this is the left Riemann sum.
$1.75(26)$ $+ 1.75(30)$ $+ 1.75(35)$ $+ 1.75(41)$ : The widths are not equal. Using 7/4 for every width ignores the table: the widths are 1, 2, 1 and 3.
$1(26) + 3(30) + 4(35) + 7(41)$ : These multipliers are the times t, not the widths of the subintervals. A width is a difference of two times.
Correct: $1(26) + 2(30) + 1(35) + 3(41)$ : Correct: each width times the rate at the right end of that subinterval.

3. Use a trapezoidal sum with the four subintervals indicated by the table to estimate the gallons of rainwater that flow into the tank from $t = 0$ to $t = 7$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$1 \cdot \frac{20 + 26}{2}$ $+ 2 \cdot \frac{26 + 30}{2}$ $+ 1 \cdot \frac{30 + 35}{2}$ $+ 3 \cdot \frac{35 + 41}{2}$ $= 23$ $+ 56$ $+ 32.5$ $+ 114$ $= 225.5$ gallons. It is also the average of the left and right sums: $\frac{207 + 244}{2}$ .

Answer: 225.5 gallons

4. Is the left Riemann sum of 207 gallons an overestimate or an underestimate of the water that flows in, and why?

An overestimate, because the rates in the table are positive
An underestimate, because R is increasing on each subinterval
An overestimate, because R is increasing on each subinterval
An underestimate, because the subintervals have unequal widths

Show the answer

Because $R$ is increasing, on each subinterval $R$ is smallest at the left end. Every left rectangle lies below the graph, so the left sum underestimates.

An overestimate, because the rates in the table are positive: Positive rates make the amount grow, but they say nothing about whether rectangles are too tall or too short.
Correct: An underestimate, because R is increasing on each subinterval: Correct: the left end has the smallest rate on each subinterval, so each rectangle is too short.
An overestimate, because R is increasing on each subinterval: For an increasing rate the left end has the smallest value on each subinterval, so the rectangles fall short.
An underestimate, because the subintervals have unequal widths: Unequal widths do not decide the direction; the monotonicity of R does.

5. For which function is a right Riemann sum on $[1, 4]$ sure to be an overestimate of the area under its graph?

a positive function that is increasing on [1, 4]
a positive function that is decreasing on [1, 4]
a positive function that is concave down on [1, 4]
a positive function with a maximum at x = 2.5

Show the answer

If $f$ is increasing, the right end of each subinterval gives the largest value of $f$ on it. Each rectangle covers its piece of the region and more, so the right sum is an overestimate.

Correct: a positive function that is increasing on [1, 4]: Correct: the right end has the largest value on each subinterval, so every rectangle is too tall.
a positive function that is decreasing on [1, 4]: Then the right end has the smallest value on each subinterval: a right sum underestimates.
a positive function that is concave down on [1, 4]: Concavity decides trapezoids and midpoints. For right sums the deciding fact is whether f increases or decreases.
a positive function with a maximum at x = 2.5: Such a function rises and then falls, so some right rectangles are too tall and some too short: no conclusion.

6. A function $h$ has $h(0) = 4$ , $h(2) = 6$ and $h(5) = 9$ . What is the trapezoidal sum for $h$ on $[0, 5]$ with the subintervals $[0, 2]$ and $[2, 5]$ ?

39
32.5
31.25
65

Show the answer

$2 \cdot \frac{4 + 6}{2}$ $+ 3 \cdot \frac{6 + 9}{2}$ $= 10$ $+ 22.5$ $= 32.5$ .

39: This is the right Riemann sum, 2(6) + 3(9).
Correct: 32.5: Correct: 2 times 5 plus 3 times 7.5.
31.25: This uses the equal-width rule with width 2.5. The subintervals here have widths 2 and 3.
65: This forgets to halve: a trapezoid uses the average of its two heights.

Part 9 · Summary

Summary

A Riemann sum estimates accumulated change by adding width times height over subintervals; a trapezoidal sum averages the two end heights. Take every width from the table. Left and right sums under- or overestimate according to whether the function increases or decreases; trapezoids and midpoints according to its concavity.

Part 10 · Up next

What comes next

Part 11 · Connections