Lesson 17
Riemann Sums and Definite Integrals
Annoucements
Quizzes
Homework
1
Today we’ll use Riemann Sums with functions that are not nice! The world is full
of functions that are just plain difficult to work with. Let’s look at some of these
We will use:
RectangleSum[<Func>, <Start x>,<End x>,<Number>,<Position>]
Position:
0 corresponds to left endpoints
0.5 corresponds to midpoints
1 corresponds to right endpoints
Example:
Approximate the area between the x-axis and the graph of
f ( x) 
15
on the interval [0.0755, 8.21] using 29 rectangles and
0.083x  19.17 x  1
2
midpoints.
Grab GGB
2
Integral symbol:
8.21

f ( x )dx
0.075
Integrating “against” the x-axis.
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Question 1
3
Sometimes you want Upper Sums or Lower Sums – these are related to Riemann
Sums, BUT different in that the calculation uses the largest y-value in each
rectangle regardless of where it occurs in the rectangle for an Upper Sum (and
similarly, the smallest for a Lower Sum).
Naturally there’s a GGB command called UpperSum and one called LowerSum!
You’ll need to provide
[<func>,<Start x>,<End x>, <Number of Rectangles>]
Example
Find the upper sum on the interval [-3, 2] using 35 rectangles for
f ( x) 
ex
3
2
Compare this value to the Riemann Sum on the same interval using left endpoints
and the same number of rectangles.
Then compute the exact value using the definite Integral command.
UpperSum:
18.94 square units
RectangleSumLeft:
18.41 square units
Definite Integral:
18.67 square units
On the interval [ −3 to 2]. The top 2 are approximates and the last is exact.
These are all AREAS between the x-axis and the curve because there are NO
ROOTS in the interval! The curve is always ABOVE the x-axis.
4
If the Riemann Sum exists on the interval [a, b], we can take a limit as the number
of rectangles approaches infinity and get the exact area. Doing this is called
“taking the definite integral from a to b”. The number a is called the lower limit of
integration, and the number b is called the upper limit of integration.
5
Properties of Definite Integrals.
Suppose f ( x) and g ( x ) are integrable function. Then we know the following:
a
1
 f ( x)dx  0
the area from a to a is zero!
a
2
b
b
a
a
 cf ( x)dx  c  f ( x)dx
this is really similar to the derivative rule for multipliers!
3
b
a
a
b
 f ( x)dx    f ( x)dx
If you exchange the limits of integration, you get “negative sign on the area”…on
the other hand, when you are integrating BELOW the x-axis, reverse the order to
get “positive area”
b
4

a
f ( x)  g ( x)dx 
a

b
a
f ( x)dx   g ( x) dx
b
Again – very similar to the property for derivatives…you can do them in pieces
5
For the interval [a, c, b] where c is an “inbetween” number
b

a
c
f ( x)dx 

a
b
f ( x)dx   f ( x)dx
c
This will come in usefully when we’ve got those zeros and the “negative sign on
the area” below the x-axis thing going on.
6
Let’s get some vocabulary and ideas fixed in place.
If the instructions ask for the “definite integral” on a closed interval, then you just
use the Integral command and take the NET as your answer.
Visually
If the instructions ask for the “area”, then you reverse the sign on the parts below
the x-axis and add all the POSITIVE numbers.
Visual:
Definite integral and area are the SAME number if the function stays above the
x-axis on the interval.
7
Let’s do two of these!
8
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Question 2
9
Now, let’s review the definite integral command:
It will give you the NET of the areas above and below the x-axis adding the areas
above and subtracting the areas below if you do not intervene. If the function is
totally non-negative in the interval in which you are integrating then just do the
Integral command.
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Question 3
10
Example:
2
Evaluate:
x
1
3
x
Let’s look at the graph:
What is the area between the curve and the x-axis?
11
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Question 4
12
Example
5
Evaluate
 3( x
2
 2)dx
1
Integrate!
Definite integral (net) = 90
True area = 111.32
Why are these different?
13
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Question 5
14
We’ll ask questions like this if we want you to note the negative area thing.
15
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Question 6
16
Popper 17 – Questions 7 through 10
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