Math lecture 10 (Introduction to Integration)

Introduction to Integration

Integration is a way of adding slices to find the whole.

Integration can be used to find areas, volumes, central points and many useful

things. But it is easiest to start with finding the area under the curve of a

function like this:

What is the area under y = f(x) ?

Slices

We could calculate the function at a few points

and add up slices of width Δx like this (but the

answer won't be very accurate):

We can make Δx a lot smaller and add up many

small slices (answer is getting better):

And as the slices approach zero in width, the

answer approaches thetrue answer.

We now write dx to mean the Δx slices are

approaching zero in width.

That is a lot of adding up!

But we don't have to add them up, as there is a "shortcut". Because ...

... finding an Integral is the reverse of finding a Derivative.

(So you should really know about Derivatives before reading more!)

Like here:

Example: What is an integral of 2x?

We know that the derivative of x2 is 2x ...

... so an integral of 2x is x2

We will show you more examples later.

Notation

The symbol for "Integral" is a

stylish "S"

(for "Sum", the idea of summing

slices):

After the Integral Symbol we put the function we want to find the integral of

(called the Integrand),

and then finish with dx to mean the slices go in the x direction (and approach

zero in width).

And here is how we write the answer:

Plus C

We wrote the answer as x2 but why + C ?

It is the "Constant of Integration". It is there because of all the functions

whose derivative is 2x:

The derivative of a constant is zero, so when we reverse the operation (to find

the integral) there could have been a constant of any value.

So we wrap up the idea by just writing + C at the end.

Tap and Tank

Integration is like filling a tank from a tap.

The input (before integration) is the flow rate from the

tap.

Integrating the flow (adding up all the little bits of

water) gives us thevolume of water in the tank.

Imagine the flow starts at 0 and gradually increases (maybe a motor is slowly

opening the tap).

As the flow rate increases, the tank

fills up faster and faster.

With a flow rate of 2x, the tank fills

up at x2.

We have integrated the flow to get

the volume.

Example: (assuming the flow is in liters per minute) after 3 minutes

(x=3):

the flow rate has reached 2x = 2×3 = 6 liters/min,

and the voume has reached x2 = 32 = 9 liters.

We can do the reverse, too:

Imagine you don't know the flow

rate.

You only know the volume is

increasing by x2.

You can go in reverse (using the

derivative, which gives us the slope)

and find that the flow rate is 2x.

Example: at 2 minutes the slope of the volume is 4, meaning it is

increasing at 4 liters/minute, which is the flow rate. Likewise at 3 minutes

the slope will be 6, etc.

So Integral and Derivative are opposites.

We can write that down this way:

The integral of the flow rate 2x tells us the volume of water: ∫2x dx =

x2 + C

And the slope of the volume increase x2+C gives us back the

flow rate: (x2 + C) =

2x

And hey, we even get a nice

explanation of that "C" value ...

maybe the tank already had water in

it!

The flow would still increase

the volume by the same

amount

And the increase in volume

can give us back the flow.

Which teaches us to always add "+

C".

Other functions

Well, we have played with y=2x enough now, so how do we integrate other

functions?

If you are lucky enough to find the function on the result side of a derivative,

then, using your knowledge that derivatives and integrals are opposites, you

have an answer (but remember to add C).

Example: what is ∫cos(x) dx ?

From the Rules of Derivatives table you see the derivative of sin(x) is cos(x) so:

∫cos(x) dx = sin(x) + C

But a lot of the "reversing" has already been done (see Rules of Integration).

Example: What is ∫x3 dx ?

On Rules of Integration there is a "Power Rule" that says:

∫xn dx = xn+1/(n+1) + C

We can use that rule with n=3:

∫x3 dx = x4 /4 + C

Knowing how to use those rules is the key to being good at

Integration.

Definite vs Indefinite Integrals

We have been doing Indefinite Integrals so far.

A Definite Integral has actual values to calculate between (they are put at the

bottom and top of the "S"):

Indefinite Integral Definite Integral

(a)

There’s not really a whole lot to do here other than use the first two formulas from the

beginning of this section. Remember that when integrating powers (that aren’t -1 of

course) we just add one onto the exponent and then divide by the new exponent.

Be careful when integrating negative exponents. Remember to add one onto the

exponent. One of the more common mistakes that students make when integrating

negative exponents is to “add one” and end up with an exponent of “-7” instead of the

correct exponent of “-5”. [Return to Problems]

(b)

This is here just to make sure we get the point about integrating negative exponents.

Integration by Substitution

"Integration by Substitution" (also called "u-substitution") is a method to find

an integral, but only when it can be set up in a special way.

The first and most vital step is to be able to write your integral in this form:

Note that you have g(x) and its derivative g'(x)

It might make more sense as is an example:

Here f=cos, and you have g=x2 and its derivative of 2x

This integral is good to go!

If your integral is set up like that, then you can do this substitution:

Then you can integrate f(u), and finish by putting g(x) back as a replacement

for u.

Like this:

Example: ∫cos(x2) 2x dx

We know (from above) that it is in the right form to do the substitution:

Now integrate:

∫cos(u) du = sin(u) + C

And finally put u=x2 back again:

sin(x2) + C

So ∫cos(x2) 2x dx = sin(x2) + C worked out really nicely! (Well, I knew it

would.)

This method only works on some integrals of course, and it may need

rearranging:

Example: ∫cos(x2) 6x dx

Oh no! It is 6x, not 2x. Our perfect setup is gone.

Never fear! Just rearrange the integral like this:

∫cos(x2) 6x dx = 3∫cos(x2) 2x dx

(We can pull constant multipliers outside the integration, see Rules of

Integration.)

Then proceed as before:

3∫cos(u) du = 3 sin(u) + C

Now put u=x2 back again:

3 sin(x2) + C

Done!

Now we are ready for a slightly harder example:

Example: ∫x/(x2+1) dx

Let me see ... the derivative of x2+1 is 2x ... so how about we rearrange it

like this:

∫x/(x2+1) dx = ½∫2x/(x2+1) dx

Then we have:

Then integrate:

½∫1/u du = ½ ln(u) + C

Now put u=x2+1 back again:

½ ln(x2+1) + C

And how about this one:

Example: ∫(x+1)3 dx

Let me see ... the derivative of x+1 is ... well it is simply 1.

So we can have this:

∫(x+1)3 dx = ∫(x+1)3 · 1 dx

Then we have:

Then integrate:

∫u3 du = (u4)/4 + C

Now put u=x+1 back again:

(x+1)4 /4 + C

So there you have it.

In Summary

If you can put an integral in this form:

Then you can make u=g(x) and integrate ∫f(u) du

And finish up by re-inserting g(x) where u is.

Integration by Parts

Integration by Parts is a special method of integration that is often useful when

two functions are multiplied together, but is also helpful in other ways.

You will see plenty of examples soon, but first let us see the rule:

∫u v dx = u∫v dx −∫u' (∫v dx) dx

u is the function u(x)

v is the function v(x)

As a diagram:

And let us get straight into an example:

Example: What is ∫x cos(x) dx ?

First choose u and v:

u = x

v = cos(x)

Differentiate u: u' = x' = 1

Integrate v: ∫v dx = ∫cos(x) dx = sin(x)

Now put it together:

Simplify and solve:

x sin(x) − ∫sin(x) dx

x sin(x) + cos(x) + C

So we followed these steps:

Choose u and v

Differentiate u: u'

Integrate v: ∫v dx

Put u, u' and ∫v dx here: u∫v dx −∫u' (∫v dx) dx

Simplify and solve

In English, to help you remember, ∫u v dx becomes:

(u integral v) minus integral of (derivative u, integral v)