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Course: Staging content lifeboat > Unit 11

Lesson 2: Integral calc staging

(Not done) Worksheet: Riemann sums with net areas

Practice Riemann sums withs sigma notations and functions that go below the x-axis.

Interesting things happen to the Riemann sum when the graph of the function goes below the

x

-axis. At these points, the value of the function is negative, so the value of the Riemann rectangles is opposite of their area!

Example

Let's try it out with

g (x) = - 2 x + 4

on the interval

[0, 5]

We can create

L (5)

as a left Riemann sum with

5

subdivisions.

From the graph, we have:

\begin{aligned} L (5) & = g (0) + g (1) + g (2) + g (3) + g (4) \\ = 4 + 2 + 0 - 2 - 4 \\ = 0 \end{aligned}

Our sum equals

0

! How did that happen? The areas of the rectangles

above the x -axis

are exactly equal to the areas of the rectangles

below the x -axis

. This means that their net areas cancel out.

Using symmetry

The graph of

f (x) = \cos \frac{π x}{2}

with sixteen midpoint rectangles is shown below.

Let

M (16)

be the midpoint Riemann sum where:

M (16) = \sum_{n = 1}^{16} \cos (x_{n}) Δ x

We see that

f (x)

has rotational symmetry about the point

(1, 0)

. The

4

rectangles to the left of this point have the same area as the

4

rectangles to the right, but the

4

on the right are below the

x

-axis. Thus:

\sum_{n = 13}^{16} \cos (x_{n}) Δ x = - \sum_{n = 9}^{12} \cos (x_{n}) Δ x

Problem 1

Which of the following sums are equal to $0$ ‍ ?

The sum

\sum_{n = 1}^{8} \cos (x_{n}) Δ x

has rotational symmetry about the point

(- 1, 0)

. The

4

rectangles to the left of this point have the same area as the

4

rectangles to the right, but the

4

on the left are below the

x

-axis, as shown in the following graph.

So:

\begin{aligned} \sum_{n = 1}^{8} \cos (x_{n}) Δ x & = \sum_{n = 1}^{4} \cos (x_{n}) Δ x + \sum_{n = 5}^{8} \cos (x_{n}) Δ x \\ = - (\sum_{n = 5}^{8} \cos (x_{n}) Δ x) + \sum_{n = 5}^{8} \cos (x_{n}) Δ x \\ = 0 \end{aligned}

The sum

\sum_{n = 5}^{16} \cos (x_{n}) Δ x

has rotational symmetry about the point

(1, 0)

, as previously noted. The graph below shows that the Riemann sum includes

8

rectangles to the left of this point and

4

to the right.

So:

\begin{aligned} \sum_{n = 5}^{16} \cos (x_{n}) Δ x & = \sum_{n = 5}^{8} \cos (x_{n}) Δ x + \sum_{n = 9}^{12} \cos (x_{n}) Δ x + \sum_{n = 13}^{16} \cos (x_{n}) Δ x \\ = \sum_{n = 5}^{8} \cos (x_{n}) Δ x + \sum_{n = 9}^{12} \cos (x_{n}) Δ x - (\sum_{n = 9}^{12} \cos (x_{n}) Δ x) \\ = \sum_{n = 5}^{8} \cos (x_{n}) Δ x \end{aligned}

Because

\sum_{n = 5}^{8} \cos (x_{n}) Δ x

includes only rectangles above the

x

-axis,

\sum_{n = 5}^{16} \cos (x_{n}) Δ x > 0

The sum

\sum_{n = 1}^{16} \cos (x_{n}) Δ x

has reflective symmetry across the

y

-axis. The sum to the left of the

y

-axis are exactly equal to sum to the right of the

y

-axis.

So:

\begin{aligned} \sum_{n = 1}^{16} \cos (x_{n}) Δ x & = 2 \cdot \sum_{n = 1}^{8} \cos (x_{n}) Δ x \\ = 2 (0) \\ = 0 \end{aligned}

Solution

The following sums are equal to

0

\sum_{n = 1}^{8} \cos (x_{n}) Δ x

\sum_{n = 1}^{16} \cos (x_{n}) Δ x

Problem 2

Match the sums that have equal values.

1


$\sum_{n = 1}^{8} h (n) Δ x$ ‍	$=$ ‍	$\sum_{n = 1}^{16} h (n) Δ x$ ‍
$\sum_{n = 5}^{8} h (n) Δ x$ ‍	$=$ ‍	$\sum_{n = 9}^{12} h (n) Δ x$ ‍
$- \sum_{n = 5}^{8} h (n) Δ x$ ‍	$=$ ‍	$\sum_{n = 1}^{4} h (n) Δ x$ ‍

Comparing sums

The graph of

f

is shown below. A total of

24

right-hand rectangles are shown, with

8

above

the

x

-axis and

16

below

the

x

-axis. All

24

of the rectangles have the same width.

Problem 3

Sort the values from least to greatest, with the least value at the top.

The sum

\sum_{i = 1}^{14} f (\frac{i}{2}) \cdot \frac{1}{2}

gives the

14

rectangles from

x = 0

x = 7

. These are all below the

x

-axis, so the sum is negative.

The sum

\sum_{i = 1}^{8} f (\frac{i}{2}) \cdot \frac{1}{2}

gives

8

of the rectangles from the previous sum. Since it leaves out some of the negative values of the previous sum, this sum is greater.

The sum

\sum_{i = 1}^{12} f (- 5 + \frac{i}{2}) \cdot \frac{1}{2}

gives

12

rectangles from

x = - 5

x = 1

, including all

8

of the rectangles above the

x

-axis, and a few negative values. The sum is positive.

The sum

\sum_{i = 1}^{8} f (- 5 + \frac{i}{2}) \cdot \frac{1}{2}

gives all

8

rectangles with positive values, but none of the negative values. This is the greatest sum.

Solution


$\sum_{i = 1}^{14} f (\frac{i}{2}) \cdot \frac{1}{2}$ ‍
$\sum_{i = 1}^{8} f (\frac{i}{2}) \cdot \frac{1}{2}$ ‍
$\sum_{i = 1}^{12} f (- 5 + \frac{i}{2}) \cdot \frac{1}{2}$ ‍
$\sum_{i = 1}^{8} f (- 5 + \frac{i}{2}) \cdot \frac{1}{2}$ ‍

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