Post on 28-Jun-2020
MAT01A1: Integration
Dr Craig
Week: 18 May 2020
The purpose of this set of slides is to
introduce and explain the topic of Chapter 5:
integration.
Before we start on that, let us re-cap the
main idea from Chapter 4.9.
Antiderivatives
A function F is an antiderivative of f on
an interval I if
F ′(x) = f (x)
for all x ∈ I .
Example: F (x) = x2 is an antiderivative of
f (x) = 2x.
Question: are antiderivatives unique?
Answer: No, antiderivatives are not unique.
Consider the previous example of f (x) = 2x
and F (x) = x2.
The functions
G(x) = x2 + 4 or H(x) = x2 + 7
are also antiderivatives of f (x) = 2x because
G′(x) = 2x and H ′(x) = 2x
Theorem: If F is an antiderivative of f
on an interval I , then the most general
antiderivative of f on I is F (x) + C
where C is an arbitrary constant.
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2 + 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2 + 4x + C
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2 + 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2 + 4x + C
Chapter 5
Area under a curve
How do we calculate the area under a curve?
This chapter is all about how we calculate the areaunder a curve. Section 5.1 covers the main idea ofhow we do this. There are a number of lengthycalculations in the Examples in 5.1. We don’t coverthose in detail. What we want is for you to have agood understanding of the principles behind thecalculation of the area.
In the second semester, we will take curves androtate them around different axes and thencalculate the volumes and surface areas of thosethree-dimensional shapes. These volume andsurface calculations rely on the same principles thatwe see here in 5.1.
Approximating the area under a curve
We can approximate the area under the
curve y = f (x) between x = a and x = b by
adding up the areas of a number of
rectangles.
Over the next few slides we
explain the diagram above.
Approximating the area under a curve
We can approximate the area under the
curve y = f (x) between x = a and x = b by
adding up the areas of a number of
rectangles. Over the next few slides we
explain the diagram above.
Approximating the area under a curve
In the diagram above, we use just four
rectangles for our approximation.
The length
of the base of each rectangle is b−a4 . The
sum of the areas of the rectangles
approximates the area under the curve.
Approximating the area under a curve
In the diagram above, we use just four
rectangles for our approximation. The length
of the base of each rectangle is b−a4 .
The
sum of the areas of the rectangles
approximates the area under the curve.
Approximating the area under a curve
In the diagram above, we use just four
rectangles for our approximation. The length
of the base of each rectangle is b−a4 . The
sum of the areas of the rectangles
approximates the area under the curve.
Approximating the area under a curve
The other thing that you will have noticed is
that there are three different approximations.
For each sketch, the y-value used for the
height of the rectangles is different. Now we
explain these different options.
Approximating the area under a curve
The other thing that you will have noticed is
that there are three different approximations.
For each sketch, the y-value used for the
height of the rectangles is different.
Now we
explain these different options.
Approximating the area under a curve
The other thing that you will have noticed is
that there are three different approximations.
For each sketch, the y-value used for the
height of the rectangles is different. Now we
explain these different options.
Approximating the area under a curve
Consider the first rectangle in each sketch.
We have different options for the height of
this rectangle. In the first sketch, h = f (a),
in the second sketch, h = f (x1), and in the
third sketch h = f(a+x12
).
Approximating the area under a curve
Consider the first rectangle in each sketch.
We have different options for the height of
this rectangle.
In the first sketch, h = f (a),
in the second sketch, h = f (x1), and in the
third sketch h = f(a+x12
).
Approximating the area under a curve
Consider the first rectangle in each sketch.
We have different options for the height of
this rectangle. In the first sketch, h = f (a),
in the second sketch, h = f (x1),
and in the
third sketch h = f(a+x12
).
Approximating the area under a curve
Consider the first rectangle in each sketch.
We have different options for the height of
this rectangle. In the first sketch, h = f (a),
in the second sketch, h = f (x1), and in the
third sketch h = f(a+x12
).
Approximating the area under a curve
The label under each picture tells us which
x-value from each interval we are using to
calculate the heights of the rectangles.
The
height of a rectangle is the y-value of the
curve at the chosen x-value.
Approximating the area under a curve
The label under each picture tells us which
x-value from each interval we are using to
calculate the heights of the rectangles. The
height of a rectangle is the y-value of the
curve at the chosen x-value.
Approximating the area under a curve
Notice: the choice of x-value changes the
approximation.
Here, if the left-end point of
the intervals is used to calculate the height,
then the sum of the four rectangles
underestimates the area under the curve.
Approximating the area under a curve
Notice: the choice of x-value changes the
approximation. Here, if the left-end point of
the intervals is used to calculate the height,
then the sum of the four rectangles
underestimates the area under the curve.
Approximating the area under a curve
Notice: the choice of x-value changes the
approximation. Here, if the right-end point
of the intervals is used to calculate the
height, then the sum of the four rectangles
overestimates the area under the curve.
Approximating the area under a curve
If we were to change from four to n
rectangles, then the length of the base of
each rectangle would be ∆x =b− an
. As we
increase n, the width of each rectangle
decreases (they become skinnier).
Now we are going to show an example in
more detail.
The screen shots come from this online tool:
Estimating areas under curves
(It is a nice tool to play around with. Click
on “Integrals” on the left-hand side and then
select one of the examples from 5.1 or 5.2.)
We are going to work with the curve
y = 4xe−x
and we are interested in the area under the
curve from x = 0 to x = 4.
Using more advanced integration techniques
(that you will learn in MAT01B1) we know
that the area is equal to 4− 20e4≈ 3.63.
We start off by using eight rectangles. We
have an approximation for each different
choice of how to calculate the height.
We are going to work with the curve
y = 4xe−x
and we are interested in the area under the
curve from x = 0 to x = 4.
Using more advanced integration techniques
(that you will learn in MAT01B1) we know
that the area is equal to 4− 20e4≈ 3.63.
We start off by using eight rectangles. We
have an approximation for each different
choice of how to calculate the height.
We are going to work with the curve
y = 4xe−x
and we are interested in the area under the
curve from x = 0 to x = 4.
Using more advanced integration techniques
(that you will learn in MAT01B1) we know
that the area is equal to 4− 20e4≈ 3.63.
We start off by using eight rectangles. We
have an approximation for each different
choice of how to calculate the height.
Now we will increase the number of
rectangles to 30 (i.e. we now divide the
interval from 0 to 4 into 30 sub-intervals
instead of just eight).
Notice how the absolute value of the error
for all three calculations decreases.
There are two very important things that we mustobserve from those six diagrams:
I The accuracy of the approximation improves asthe number of rectangles increases.
I The difference between each of the threemethods decreases as the number of rectanglesincreases.
The first point might be expected, but the secondone might be surprising. Why does this happen?Basically, because the length of each sub-interval(i.e. the base of each rectangle) is getting shorter,the difference between the left, right and middle ofthe interval becomes less influential.
There are two very important things that we mustobserve from those six diagrams:
I The accuracy of the approximation improves asthe number of rectangles increases.
I The difference between each of the threemethods decreases as the number of rectanglesincreases.
The first point might be expected, but the secondone might be surprising. Why does this happen?Basically, because the length of each sub-interval(i.e. the base of each rectangle) is getting shorter,the difference between the left, right and middle ofthe interval becomes less influential.
There are two very important things that we mustobserve from those six diagrams:
I The accuracy of the approximation improves asthe number of rectangles increases.
I The difference between each of the threemethods decreases as the number of rectanglesincreases.
The first point might be expected, but the secondone might be surprising. Why does this happen?Basically, because the length of each sub-interval(i.e. the base of each rectangle) is getting shorter,the difference between the left, right and middle ofthe interval becomes less influential.
A formula for the area under a curve
What we have seen is that as the intervals
get smaller (i.e. when n increases), it doesn’t
matter which x-value in the interval we use
to find the height of a rectangle.
So, let x∗i be any x-value in the i-th interval.
(These are called the sample points.) Recall
that the length of an interval is ∆x = b−an .
Now, the area under y = f (x) from a to b is:
A = limn→∞
(f (x∗1)∆x+f (x∗2)∆x+· · ·+f (x∗n)∆x
)
A formula for the area under a curve
What we have seen is that as the intervals
get smaller (i.e. when n increases), it doesn’t
matter which x-value in the interval we use
to find the height of a rectangle.
So, let x∗i be any x-value in the i-th interval.
(These are called the sample points.) Recall
that the length of an interval is ∆x = b−an .
Now, the area under y = f (x) from a to b is:
A = limn→∞
(f (x∗1)∆x+f (x∗2)∆x+· · ·+f (x∗n)∆x
)
A formula for the area under a curve
What we have seen is that as the intervals
get smaller (i.e. when n increases), it doesn’t
matter which x-value in the interval we use
to find the height of a rectangle.
So, let x∗i be any x-value in the i-th interval.
(These are called the sample points.) Recall
that the length of an interval is ∆x = b−an .
Now, the area under y = f (x) from a to b is:
A = limn→∞
(f (x∗1)∆x+f (x∗2)∆x+· · ·+f (x∗n)∆x
)
A formula for the area under a curve
Let A be the area under the curve y = f (x)
from x = a to x = b. We have ∆x =b− an
.
Then
A = limn→∞
(f (x∗1)∆x+f (x∗2)∆x+· · ·+f (x∗n)∆x
)
or, in sigma notation:
A= limn→∞
( n∑i=1
f (x∗i )∆x)
.
Notice that as n→∞ the number of
rectangles increases, but ∆x→ 0.
A formula for the area under a curve
Let A be the area under the curve y = f (x)
from x = a to x = b. We have ∆x =b− an
.
Then
A = limn→∞
(f (x∗1)∆x+f (x∗2)∆x+· · ·+f (x∗n)∆x
)or, in sigma notation:
A= limn→∞
( n∑i=1
f (x∗i )∆x)
.
Notice that as n→∞ the number of
rectangles increases, but ∆x→ 0.
Take a short break here before we start with
Chapter 5.2.
Let f be defined for a 6 x 6 b and let
∆x =b− an
. Let x∗i be a sample point in
the i-th subinterval.
The definite integral of f from a to b is∫ b
a
f (x) dx = limn→∞
( n∑i=1
f (x∗i ).∆x)
The definite integral of f from a to b exists
provided that the limit on the RHS exists. If
the limit (and hence the integral) exists, we
say that f is integrable on [a, b].
Let f be defined for a 6 x 6 b and let
∆x =b− an
. Let x∗i be a sample point in
the i-th subinterval.
The definite integral of f from a to b is∫ b
a
f (x) dx = limn→∞
( n∑i=1
f (x∗i ).∆x)
The definite integral of f from a to b exists
provided that the limit on the RHS exists. If
the limit (and hence the integral) exists, we
say that f is integrable on [a, b].
Let f be defined for a 6 x 6 b and let
∆x =b− an
. Let x∗i be a sample point in
the i-th subinterval.
The definite integral of f from a to b is∫ b
a
f (x) dx = limn→∞
( n∑i=1
f (x∗i ).∆x)
The definite integral of f from a to b exists
provided that the limit on the RHS exists. If
the limit (and hence the integral) exists, we
say that f is integrable on [a, b].
Some points about notation∫ b
a
f (x) dx = limn→∞
( n∑i=1
f (x∗i )∆x)
I the∫
symbol is like a stretched out S (it
was introduced by the mathematician
Gottfried Leibniz because an integral is a
limit of sums);
I the sumn∑i=1
f (x∗i )∆x is known as a
Riemann sum after the mathematician
Bernhard Riemann.
Some more points about notation∫ b
a
f (x) dx
What the notation above represents is: the
area between the curve y = f (x) and the
x-axis from x = a to x = b.
I a < b;
I f (x) is called the integrand, a is the
lower limit and b is the upper limit;
I The procedure of calculating an integral
is called integration.
Some more points about notation∫ b
a
f (x) dx
What the notation above represents is: the
area between the curve y = f (x) and the
x-axis from x = a to x = b.
I a < b;
I f (x) is called the integrand, a is the
lower limit and b is the upper limit;
I The procedure of calculating an integral
is called integration.
Some more points about notation∫ b
a
f (x) dx
I the dx at the end is important. It
indicates that we are integrating with
respect to x.
I This integral exists if a limit exists, in
which case it is just a real number.
I You could also write the integral as∫ ba f (t) dt or
∫ ba f (z) dz. (Notice how the
variable changes in both places.)
What if the curve has −ve y-values?
The red region has area 4. The purple region
has area 2, but this is treated as negative
because it is below the x-axis.
What if the curve has −ve y-values?
We get∫ π/2−π 2 cosx dx =∫ −π/2
−π2 cosx dx+
∫ π/2
−π/22 cosxdx = −2+4 = 2
Chapter 2.5 was all about continuity. We
found that most of the functions we work
with are continuous everywhere on their
domains.
This is very useful because it turns
out that continuous functions are very well
behaved with respect to integration.
Theorem: If f is continuous on [a, b], or
if f has only a finite number of jump
discontinuities, then f is integrable on
[a, b]; that is, the definite integral∫ ba f (x) dx exists.
Chapter 2.5 was all about continuity. We
found that most of the functions we work
with are continuous everywhere on their
domains. This is very useful because it turns
out that continuous functions are very well
behaved with respect to integration.
Theorem: If f is continuous on [a, b], or
if f has only a finite number of jump
discontinuities, then f is integrable on
[a, b]; that is, the definite integral∫ ba f (x) dx exists.
Chapter 2.5 was all about continuity. We
found that most of the functions we work
with are continuous everywhere on their
domains. This is very useful because it turns
out that continuous functions are very well
behaved with respect to integration.
Theorem: If f is continuous on [a, b], or
if f has only a finite number of jump
discontinuities, then f is integrable on
[a, b]; that is, the definite integral∫ ba f (x) dx exists.
After the theorem on the previous slide the
textbook looks at integrals and their
associated Riemann sums in a lot of detail.
We briefly look at Example 4 before moving
on to properties of the integral.
Examples:
Evaluate the integrals by interpreting them in
terms of area:
1.
∫ 1
0
√1− x2 dx
2.
∫ 3
0
(x− 1) dx
Solutions: Use the sketches below to find the
solutions and then check your answers.
1.∫ 1
0
√1− x2 dx = π
4
2.∫ 3
0 (x− 1) dx = 1.5
Solutions: Use the sketches below to find the
solutions and then check your answers.
1.∫ 1
0
√1− x2 dx = π
4
2.∫ 3
0 (x− 1) dx = 1.5
Properties of the definite integral
Earlier when talking about∫ ba f (x) dx, we
assumed earlier that a < b because we were
thinking of the integral as the area under a
curve. If we think of∫ ba f (x) dx purely as
the limit-based definition then we get:∫ a
b
f (x) dx = −∫ b
a
f (x) dx
and ∫ a
a
f (x) dx = 0
Properties of the Integral
We assume f and g are continuous. Then
1.
∫ b
a
c dx = c(b− a) for c any constant
2.
∫ b
a
[f(x)+ g(x)
]dx =
∫ b
a
f(x) dx+
∫ b
a
g(x) dx
3.
∫ b
a
cf(x) dx = c
∫ b
a
f(x) dx for c any constant
4.
∫ b
a
[f(x)− g(x)
]dx =
∫ b
a
f(x) dx−∫ b
a
g(x) dx
5.
∫ c
a
f(x) dx+
∫ b
c
f(x) dx =
∫ b
a
f(x) dx
1.
∫ b
a
c dx = c(b− a) for c any constant
Property 1. follows from the fact that the
area under a constant function f (x) = c is
simply the area of a rectangle with height c
and base b− a.
2.
∫ b
a
[f (x) + g(x)
]dx =
∫ b
a
f (x) dx +
∫ b
a
g(x) dx
Property 2. can be understood by looking at
the following diagram.
5.
∫ c
a
f (x) dx +
∫ b
c
f (x) dx =
∫ b
a
f (x) dx
Property 5. says that we can break an area
into two pieces at some intermediate x value.
(This will be useful for integrating piecewise
defined functions!)
Comparison Properties of the Integral
6. If f (x) > 0 for a 6 x 6 b, then∫ b
a
f (x) dx > 0
7. If f (x) > g(x) for a 6 x 6 b, then∫ b
a
f (x) dx >∫ b
a
g(x) dx
8. If m 6 f (x) 6M for a 6 x 6 b, then
m(b− a) 6∫ b
a
f (x) dx 6 M(b− a)
Property 6 says that if the y-values of the
function f are all non-negative on the
interval [a, b], then you are guaranteed that
the integral of f over that interval will be
non-negative.
Property 7 says that if one function has
larger (or equal) y-values than another
function on [a, b], then the integral of the
function with the larger y-values will be
greater than (or equal to) the integral of the
other function.
8. If m 6 f (x) 6M for a 6 x 6 b, then
m(b− a) 6∫ b
a
f (x) dx 6 M(b− a)
Property 8. can be easily understood by the
following diagram.
Examples:
1. If∫ 5
1 f (x) dx = 12 and∫ 5
4 f (x) dx = 3.6,
find∫ 4
1 f (x) dx.
2. Use Comparison Property 8 to estimate
the value of
∫ 4
1
√x dx.
3. Calculate
∫ 2
−1|x| dx.
Solutions
1. Using Property 5. we see that∫ 4
1 f (x) dx +∫ 5
4 f (x) dx =∫ 5
1 f (x) dx so∫ 4
1 f (x) dx = 12− 3.6 = 8.4.
2. We know that for 1 6 x 6 4 we have
1 6√x 6 2. Therefore by property 8.
we have
1(4− 1) 6∫ 4
1
√x dx 6 2(4− 1)
∴ 3 6∫ 4
1
√x dx 6 6
Solutions
1. Using Property 5. we see that∫ 4
1 f (x) dx +∫ 5
4 f (x) dx =∫ 5
1 f (x) dx so∫ 4
1 f (x) dx = 12− 3.6 = 8.4.
2. We know that for 1 6 x 6 4 we have
1 6√x 6 2.
Therefore by property 8.
we have
1(4− 1) 6∫ 4
1
√x dx 6 2(4− 1)
∴ 3 6∫ 4
1
√x dx 6 6
Solutions
1. Using Property 5. we see that∫ 4
1 f (x) dx +∫ 5
4 f (x) dx =∫ 5
1 f (x) dx so∫ 4
1 f (x) dx = 12− 3.6 = 8.4.
2. We know that for 1 6 x 6 4 we have
1 6√x 6 2. Therefore by property 8.
we have
1(4− 1) 6∫ 4
1
√x dx 6 2(4− 1)
∴ 3 6∫ 4
1
√x dx 6 6
Solutions continued . . .
3. First let us sketch the area we are trying
to calculate.
Now we calculate the area of the two
triangles: 12bh = 1
2(1)(1) = 12 and
12(2)(2) = 2. So
∫ 2
−1 |x| dx = 12 + 2 = 5
2.
Solutions continued . . .
3. First let us sketch the area we are trying
to calculate.
Now we calculate the area of the two
triangles: 12bh = 1
2(1)(1) = 12 and
12(2)(2) = 2. So
∫ 2
−1 |x| dx = 12 + 2 = 5
2.
Question: How can we make definite
integrals easier to calculate?
Answer: With antiderivatives!
Prescribed tut problems:
Complete the following exercises from the
8th edition:
Ch 5.2:
1, 17, 21, 27, 29, 35, 40, 41, 42, 43, 47, 61