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