00:01
So now, it’s your chance to try
some integration with substitution.
00:05
You may be thinking that this is
getting fairly complicated calculus.
00:10
But remember that you are developing
extremely strong analytical skills.
00:14
You are developing extremely
strong problem solving skills,
which as medical students, you
will need all through your course.
00:21
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.
00:33
So, let’s now try and have
a go at this exercise.
00:37
I’ll give you some time to do this and
then we’ll go through this together.
00:41
Let’s look at our first example
that we set for you to do.
00:45
We have then an example
fairly similar to this,
but we’re asking you now
to use substitution.
00:51
So, the first example states that you
have x that’s been multiplying with
2x squared minus 5 to
the power of 3 dx.
01:03
If we use substitution,
go for the more complicated part of this
integral and make a substitution for that.
01:10
So you can say here, let u
equals to 2x squared minus 5.
01:16
So just the inside part here.
01:19
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.
01:30
Remember the second
step from the example.
01:32
You now want to change your dx.
01:35
So, when we change our dx, we
need to do du by dx equals to 4x.
01:41
Rearrange to get dx by itself.
01:43
So, du over 4x equals to dx.
01:48
Let’s replace our dx.
01:49
We’ve got x multiplied by u cube,
and then you have du over 4x.
01:56
So from here, and good things are
happening, the x is cancelled out.
02:02
Leaving you with a nice, easy integral,
we have u cubed over 4
and the integral, du.
02:10
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.
02:17
So, we have u cubed du.
02:20
nd now when we integrate, we have no
limits to change, so that’s good news.
02:24
We have 1 over 4.
02:26
Add 1 to the power,
and add 1 to the power, I mean,
u to the 4 divided by 4.
02:34
You can rewrite this
as u to the 4 over 16.
02:38
Very last bit, and I remember,
that the question asks you the
question in terms of x’s.
02:44
We can now change this
in terms of x’s again.
02:48
So we said that u was
2x squared minus 5.
02:51
So we can rewrite this u as 2x squared
minus 5 to the power of 4 over 16 plus c.
03:01
Let’s now look at our next example.
03:03
You can see that it looks
fairly complicated.
03:05
We have cos and sine and
we have square roots.
03:08
But let’s see if we can try
and make it easier now
by using some sensible
substitutions.
03:14
So the question is asking us to find
square root of 5 cos(x) minus 1
multiplied by sine(x)dx.
03:26
Now, I sense that we’ll be doing a lot
of trig differentiation and integration.
03:31
So let’s just write
that out here.
03:33
Sine goes to cos,
differentiates to cos, and
cos differentiates to -sine.
03:39
And similarly, if you want to integrate,
cos integrates to sine when I go in that
direction, and sine integrates to -cos.
03:49
Right.
03:49
I’m going to try and do this
in a fewer steps than before.
03:53
So, let’s make this
our substitution.
03:56
So let’s say let u equals
to 5cos(x) minus 1.
04:01
I know that I’m going to have to
differentiate it so I may as well do it now.
04:04
So, du by dx.
04:06
Now, look at cos.
04:08
Cos is differentiating.
04:09
So, differentiation in that direction.
04:11
Cos is differentiating to -sine.
04:14
So this becomes -5sine(x).
04:18
Rearrange to get dx by itself.
04:20
And it’s only because we know
we’re going to do that, anyway.
04:25
So, we’re kind of working a
little bit ahead of ourselves.
04:29
Now, come back to our integral.
04:31
So our integral is saying that we
have square root of 5cos(x) minus 1.
04:35
I said that to be u just to
make an easier substitution.
04:40
We’ve still got the sine(x) there,
and the dx becomes du
over -5sine(x).
04:52
So, some good things are happening here.
04:54
You can see that you have
a sine(x) and a sine(x).
04:58
So that sine(x) cancels with that.
05:01
You have this -5 here.
05:03
So, you can rewrite this as
square root of u over -5 du.
05:09
Or alternatively, you
can take the -5 outside
like this to make it easier to comprehend.
05:17
So we’ve got -1 over 5 out of the integral
and we’re integrating square root of u du.
05:23
Before we integrate, we can
change the u to u to the half du.
05:30
Remember that to integrate u, add 1
to the power, divided by new power.
05:35
So I have 1 over 5.
05:36
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.
05:44
We can tidy this up a little
bit so it becomes 1 over 5.
05:48
Take the two up from here.
05:50
We’ve discussed why
we do this earlier.
05:52
So that becomes 2u to
the 3 over 2 over 3.
05:58
And lastly, the question
was asked in terms of x’s.
06:04
So you can change
your u back to x.
06:07
So I can rewrite
this as -1 over --
multiply these together -- 15.
06:12
You still have the 2 at the top,
so the 2 can come out here.
06:15
And then the u value, we
can change back to our x.
06:19
So that was 5 cos(x) minus 1 to
the power of 3 over 2 plus c.
06:26
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.
06:43
And the last question in your exercise,
which wants you to do substitution.
06:47
Let’s have a go.
06:51
We have x squared over 2
minus x cubed squared dx.
06:57
You’ll see that this works a little bit
different to the others that we’ve done
because you have a denominator.
07:03
So you have this function
here at the bottom.
07:06
Now, remember what we could do and
what used to do in differentiation.
07:09
If there’s anything like that,
you can always take it up.
07:12
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.
07:25
Now, we can make a substitution to
make things a little bit easier.
07:29
So let’s go through this part here
and say that let u equals
to 2 minus x cubed.
07:36
If I do everything else here,
so find dy by dx as well so I can
do du by dx equals to -3x squared.
07:45
And if I rearranged to get my dx,
so that gives me du over
-3x squared equals to dx.
07:53
Now, when I put this all
in into my integral,
you’ll see that it becomes a little
bit easier than it was before.
07:59
So I have x squared.
08:00
Instead of 2 minus x cube, I can
rewrite this as u to the minus 2.
08:05
And instead of the dx, I can
rewrite du over minus 3 x squared.
08:12
It looks a little bit scary, but
when you cancel the x squares,
you see that it becomes a nice
and straightforward integral.
08:19
If I rewrite this now,
I’ve got u to the minus 2.
08:23
Let’s not forget the
-3 at the bottom.
08:25
I have du and then I have the integral 01.
08:29
So remember that this
-3 can go anywhere?
You can put it under here or you
can take it out of the integral.
08:34
It really is up to you.
08:37
The next thing to do is
to change these limits.
08:39
So, don’t forget that
these are your x limits.
08:42
So in order for this to
go into a u function,
you need to change
them to your u limits.
08:47
And for that, we can use
this function here.
08:51
So we can say that when x equals to 1,
u equals to 2 minus 1 cubed,
which just gives you 1.
08:59
And when x equals to 0, u equals to 2
minus 0 to give you an answer of two.
09:06
So we can now change our limits.
09:08
Remember that these are your u limits here.
09:11
So we’ve changed them to u's, which makes
this entire integral in terms of u’s.
09:15
Now, I’ll take this -3 out, so we can
write it as -1 over 3 on the outside.
09:20
My new u limits, when I put 1 in, I get 1.
09:23
When I put x equals to 0
in, I got a u limit of 2.
09:27
I still have u to the -2du.
09:31
And now, we integrate.
09:32
So it’s fairly straightforward
from here minus 1 over 3.
09:36
When you integrate, add 1 to the
power, divided by new power,
where limits -- we’ve
got our limits 2 and 1.
09:44
I’m just going to neaten
this up a little bit.
09:46
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.
09:55
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.
10:01
So the top limit goes in first.
10:05
So, my top limit is -1 over 1.
10:08
So the u changes to 1.
10:10
And then you minus the second
part, which is -1 over 2.
10:15
So notice what I’ve done.
10:16
I’ve just changed this u
first to 1 and then to 2.
10:20
So here, I put 1 in,
here I put 1 in, and
then I put 2 in.
10:24
And I’ve just put brackets around this
so that I don’t forget
that there’s two minuses.
10:30
Almost there.
10:31
I’ve got -1 over 3.
10:33
I’ve got -1 as over 1
over 1 as -1 plus a half.
10:37
So this minus and a
minus becomes positive.
10:41
And then I have -1 over 3.
10:43
That gives me -1 a half on the inside.
10:46
Multiply them together
to give you 1 over 6.
10:49
So the area underneath x squared
over 2 minus x cubed, all squared,
fairly complicated graph.
10:57
Between the limits 0 and 1
is 1 over 6 unit squared.