So now, it’s your chance to try
some integration with substitution.
You may be thinking that this is
getting fairly complicated calculus.
But remember that you are developing
extremely strong analytical skills.
You are developing extremely
strong problem solving skills,
which as medical students, you
will need all through your course.
You are looking at finding
areas under curves,
and although these curves look very
complicated and very mathematical,
this is a skill that you will be able to use
at any point within your medical field.
So, let’s now try and have
a go at this exercise.
I’ll give you some time to do this and
then we’ll go through this together.
Let’s look at our first example
that we set for you to do.
We have then an example
fairly similar to this,
but we’re asking you now
to use substitution.
So, the first example states that you
have x that’s been multiplying with
2x squared minus 5 to
the power of 3 dx.
If we use substitution,
go for the more complicated part of this
integral and make a substitution for that.
So you can say here, let u
equals to 2x squared minus 5.
So just the inside part here.
We can now replace this with x,
and then that becomes u to the power of 3,
because I’m saying that all of this is u,
and then I still
have dx at the end.
Remember the second
step from the example.
You now want to change your dx.
So, when we change our dx, we
need to do du by dx equals to 4x.
Rearrange to get dx by itself.
So, du over 4x equals to dx.
Let’s replace our dx.
We’ve got x multiplied by u cube,
and then you have du over 4x.
So from here, and good things are
happening, the x is cancelled out.
Leaving you with a nice, easy integral,
we have u cubed over 4
and the integral, du.
Now, the 4 at the bottom of that
confuses you, it’s just the number,
you can take it out as a fraction
outside of the integral.
So, we have u cubed du.
nd now when we integrate, we have no
limits to change, so that’s good news.
We have 1 over 4.
Add 1 to the power,
and add 1 to the power, I mean,
u to the 4 divided by 4.
You can rewrite this
as u to the 4 over 16.
Very last bit, and I remember,
that the question asks you the
question in terms of x’s.
We can now change this
in terms of x’s again.
So we said that u was
2x squared minus 5.
So we can rewrite this u as 2x squared
minus 5 to the power of 4 over 16 plus c.
Let’s now look at our next example.
You can see that it looks
We have cos and sine and
we have square roots.
But let’s see if we can try
and make it easier now
by using some sensible
So the question is asking us to find
square root of 5 cos(x) minus 1
multiplied by sine(x)dx.
Now, I sense that we’ll be doing a lot
of trig differentiation and integration.
So let’s just write
that out here.
Sine goes to cos,
differentiates to cos, and
cos differentiates to -sine.
And similarly, if you want to integrate,
cos integrates to sine when I go in that
direction, and sine integrates to -cos.
I’m going to try and do this
in a fewer steps than before.
So, let’s make this
So let’s say let u equals
to 5cos(x) minus 1.
I know that I’m going to have to
differentiate it so I may as well do it now.
So, du by dx.
Now, look at cos.
Cos is differentiating.
So, differentiation in that direction.
Cos is differentiating to -sine.
So this becomes -5sine(x).
Rearrange to get dx by itself.
And it’s only because we know
we’re going to do that, anyway.
So, we’re kind of working a
little bit ahead of ourselves.
Now, come back to our integral.
So our integral is saying that we
have square root of 5cos(x) minus 1.
I said that to be u just to
make an easier substitution.
We’ve still got the sine(x) there,
and the dx becomes du
So, some good things are happening here.
You can see that you have
a sine(x) and a sine(x).
So that sine(x) cancels with that.
You have this -5 here.
So, you can rewrite this as
square root of u over -5 du.
Or alternatively, you
can take the -5 outside
like this to make it easier to comprehend.
So we’ve got -1 over 5 out of the integral
and we’re integrating square root of u du.
Before we integrate, we can
change the u to u to the half du.
Remember that to integrate u, add 1
to the power, divided by new power.
So I have 1 over 5.
When you add 1 to the power,
that goes to u to the 3 over 2,
and then you divide
it by 3 over 2.
We can tidy this up a little
bit so it becomes 1 over 5.
Take the two up from here.
We’ve discussed why
we do this earlier.
So that becomes 2u to
the 3 over 2 over 3.
And lastly, the question
was asked in terms of x’s.
So you can change
your u back to x.
So I can rewrite
this as -1 over --
multiply these together -- 15.
You still have the 2 at the top,
so the 2 can come out here.
And then the u value, we
can change back to our x.
So that was 5 cos(x) minus 1 to
the power of 3 over 2 plus c.
So you can see that this
looked fairly complicated
but we’ve made it a little
bit easier by substitution,
and the integral at the end
was fairly straightforward
just by adding 1 to the power
and dividing by the new power.
And the last question in your exercise,
which wants you to do substitution.
Let’s have a go.
We have x squared over 2
minus x cubed squared dx.
You’ll see that this works a little bit
different to the others that we’ve done
because you have a denominator.
So you have this function
here at the bottom.
Now, remember what we could do and
what used to do in differentiation.
If there’s anything like that,
you can always take it up.
So we can rewrite this
as 0 to 1 x squared,
and you can bring the entire bracket up
and rewrite it as 2 minus x
cubed to the minus 2 dx.
Now, we can make a substitution to
make things a little bit easier.
So let’s go through this part here
and say that let u equals
to 2 minus x cubed.
If I do everything else here,
so find dy by dx as well so I can
do du by dx equals to -3x squared.
And if I rearranged to get my dx,
so that gives me du over
-3x squared equals to dx.
Now, when I put this all
in into my integral,
you’ll see that it becomes a little
bit easier than it was before.
So I have x squared.
Instead of 2 minus x cube, I can
rewrite this as u to the minus 2.
And instead of the dx, I can
rewrite du over minus 3 x squared.
It looks a little bit scary, but
when you cancel the x squares,
you see that it becomes a nice
and straightforward integral.
If I rewrite this now,
I’ve got u to the minus 2.
Let’s not forget the
-3 at the bottom.
I have du and then I have the integral 01.
So remember that this
-3 can go anywhere?
You can put it under here or you
can take it out of the integral.
It really is up to you.
The next thing to do is
to change these limits.
So, don’t forget that
these are your x limits.
So in order for this to
go into a u function,
you need to change
them to your u limits.
And for that, we can use
this function here.
So we can say that when x equals to 1,
u equals to 2 minus 1 cubed,
which just gives you 1.
And when x equals to 0, u equals to 2
minus 0 to give you an answer of two.
So we can now change our limits.
Remember that these are your u limits here.
So we’ve changed them to u's, which makes
this entire integral in terms of u’s.
Now, I’ll take this -3 out, so we can
write it as -1 over 3 on the outside.
My new u limits, when I put 1 in, I get 1.
When I put x equals to 0
in, I got a u limit of 2.
I still have u to the -2du.
And now, we integrate.
So it’s fairly straightforward
from here minus 1 over 3.
When you integrate, add 1 to the
power, divided by new power,
where limits -- we’ve
got our limits 2 and 1.
I’m just going to neaten
this up a little bit.
So I have 1 over 3, and then I’ve got
-1 over u when I bring the u down,
and the minus is just
making everything negative.
And then I’ve got 2 and 1 as my limits
Last step, we just have to put our
limits in and get our final answer.
So the top limit goes in first.
So, my top limit is -1 over 1.
So the u changes to 1.
And then you minus the second
part, which is -1 over 2.
So notice what I’ve done.
I’ve just changed this u
first to 1 and then to 2.
So here, I put 1 in,
here I put 1 in, and
then I put 2 in.
And I’ve just put brackets around this
so that I don’t forget
that there’s two minuses.
I’ve got -1 over 3.
I’ve got -1 as over 1
over 1 as -1 plus a half.
So this minus and a
minus becomes positive.
And then I have -1 over 3.
That gives me -1 a half on the inside.
Multiply them together
to give you 1 over 6.
So the area underneath x squared
over 2 minus x cubed, all squared,
fairly complicated graph.
Between the limits 0 and 1
is 1 over 6 unit squared.