Let’s have a look at our
next set of examples.
Now, I’ve put them together here because I
want you to try them out before we do them.
So take a moment to just try these out.
Remember that the last two, we
haven’t done yet as an example.
But I’m just going to give you a
little hint of thinking about ln,s.
Remember that when you integrate
1 over x, you get ln of x.
So try and relate that to these questions
and see if you can figure
this out for yourselves.
So, let’s try our first
example in the exercise.
We have the integral
of 5e to the 3x-1dx.
This is very similar to one of the examples
that we did together in the lecture.
Remember that when you integrate, the 5
stays as it is, so that doesn’t bother us.
This is a big function
and a little function,
but don’t forget, you are integrating it.
When you integrate the big
function, it doesn’t change,
so that stays as e to the 3x minus 1.
But remember to divide
it by the differential.
So the differential of this is just 3.
So we can rewrite this as 5
over 3, e to the 3x minus 1.
Nothing really simplifies,
so we put a + c at the end.
Here is your little challenge.
We are now trying to integrate
1 over 1 plus x dx.
Let’s see the kind of
things you could have done
and let’s see why we
shouldn’t have done them.
So, let’s have a look at this.
I’ve got one over 1 plus x dx.
Now, when we were differentiating
things like this,
we often brought the entire
function up to the top.
So I could rewrite this,
and I’m going to tell you
now that isn’t correct,
but I’m just showing you why this isn’t.
And remember now that if I added 1 to
the power and divided it by new power.
So imagine if I added 1 to the
power, 1 plus x minus 1 plus 1,
I get an answer of 1 plus x to
the power of 0, which is just 1.
Which is basically saying that the area
under this curve is 1 all the time,
and that’s not true.
You can change the area.
You can take smaller areas and bigger
areas with the limits that you put in.
However, this gives us no chance
to put any values in
to find the areas.
So this is obviously flawed.
We need to think of a different method.
Let’s just cross this out.
Now, I’m going to take
you back to remembering
what the differential
of ln of x was.
So, if we just remind
ourselves that ln of x,
when you differentiated it, dy by dx,
so if I put y before
this, gave you 1 over x.
So when we were introducing
all these identities,
we said that the integral
of 1 over x dx was ln of x.
Now, compare that with what you have here.
Instead of being just
x, you have 1 plus x.
So, look at this carefully for a moment.
You’ll see that if you have
1 over x goes to ln of x,
1 over 1 plus x to the power of 1, just
the way we have a power of 1 here,
must go to something that has ln in it.
So I’m going to give you a rule that’s
going to help you integrate this.
The rule states this.
If you have an integral
with f of x at the bottom, and
remember this is a linear f of x,
that means it’s f of
x to the power of 1.
And if you notice that the differential
of this function is at the top,
the answer to this is just ln of
the function at the bottom plus c.
Now, I’ve put the absolute value
there to keep it positive,
which we do when we’re dealing
with ln's because ln's is only --
ln graph is only true for
all positive values of x.
But once again, let me just remind you,
just get rid of that power so it
doesn’t look like it’s a differential.
If you have the integral and you can see
that there’s a function at the bottom,
that doesn’t have a power so
you don’t want to move it up
because otherwise, it will cancel
out like I showed you earlier.
If you have a function at the
bottom and then you can see
that the differential of
this function is at the top,
the answer to that is just
ln of that function plus c.
So, let’s get rid of this power so
it doesn’t look very confusing.
That was 1 plus x.
And let’s see if we can
apply that rule here.
The first thing that you need to spot
here is that this is going to integrate
to an ln function because
we can’t take this up.
If I did take it up, remember
what I showed you earlier,
that the powers cancel out and
you just get an answer of 1.
So, we don’t do that.
You just recognize that this
must be an ln function.
The next thing, look at the
number that you have at the top.
So firstly, the number that
we have at the top is 1.
Look at this function here, 1 plus x,
what is the
differential of 1 plus x?
So the differential
of 1 plus x is 1.
So, you are actually agreeing or you’re
actually satisfying this rule here.
That if you have a function at the bottom,
in our case, that’s 1 plus x,
the differential of that
is at the top, which is 1.
So the answer to this
must be ln of 1 plus x plus c.
So, it’s just ln of this function at
the bottom because the integral --
because the differential at the top is the
differential of the function at the bottom.
This is going to take a
little bit more practice,
and then we’ll see that we’ll
start to get better at this.
So, our next example for ln's.
My function here is 7 over 3 plus 8x dx.
First thing, remember, you may
be tempted to take 3 plus 8x up
and change its power.
But remember what happened
when we did that last time?
The power is cancelled out to
give you to the power of 0,
so the entire area became 1.
And remember that that is flawed because
that can’t be the area everywhere.
The area could be bigger or smaller.
So, what we need to do firstly is to
recognize that this must be an ln function.
The definition for that was this.
If I have a function f of x at the bottom
and I see the differential
of the function at the top,
my answer is ln of
that function plus c.
I have a function
here at the bottom.
What is the differential
of this, 3 plus 8x?
The differential of that is 8.
Do I have 8 at the top?
I have 7.
So there’s a couple of
things that I need to do.
I can tweak this equation.
Firstly, I’m allowed to move
the 7 out of the integral.
It’s just the number, it’s just
the constant, so I’ll take it out
and I still got 3 plus
8x dx on the inside.
So the 7 is just the constant,
it’s not going to be integrated.
You can take it out of the integral just
to make our lives a little bit easier.
Okay, so what do we want now?
For this to be an ln function, 3 plus 8x,
the differential of this is
8, so we want 8 at the top.
Let me just show you how I’m
tweaking this in my mind.
I want 8 to be at the top.
If that happens, then my answer
will give me the integral,
will satisfy this equation here.
However, the 8, I put in myself.
So because I tweaked it myself, I have
to make up for it; I can’t just cheat.
So because I put that 8 there,
I’m going to divide it by an 8 on
the outside just to make it fair.
So because I multiplied it with
an 8, I can divide it with an 8;
8 divided by 8 is just 1, so
I’m not changing the equation.
Now, let’s look at our
equation or our integral.
Just look here on the inside.
I have a function, 3 plus 8x,
and I have the integral of
that at the top, which is 8.
So, that’s all good news.
That means that this entire thing
can integrate to ln of 3 plus 8x.
So, that’s good news.
But don’t forget that you have
this little value on the outside.
So we’re going to put 7
over 8 on the outside
because that’s the little
tweaking that we did.
And then you put a plus c at the end.
And that is the answer to this integral.
It’s one of the harder ones and it does
require you to recognize it first.
And that’s why it’s so important to
remember all your different rules
and all your different types of functions
so you can keep remembering
them as you go through it.
If you notice that there is a function
at the bottom of an integral,
and that’s all to the power of
1 or it doesn’t have a power,
don’t be tempted to move it all
up because it will cancel out.
But instead, try and integrate it
or tweak it into an ln function
because the answer will be an ln function
but you just need to make sure
that you put the differential
of that function at the top.
If it’s not there at the top, you
can put it there yourselves,
but as long as you make up for it.
Any extra bits that you multiply,
divide it on the outside
so you make up for the little tweaks
that you’ve made to the equation.