We are now given a differential equation.
So we've been told dY by dX
equals to an expression,
and we're being asked to find Y.
So the question you're asking yourself
is, where did this dY by dX come from?
What was before this in order
for us to get this result?
And that's basically what
integration is doing.
So let's have a look
at our question.
We have here dY by d Xequals
to X squared plus 3x minus 1.
Now, to go back, we want to go
back to the original function.
And so this must have come
from an original function
and we're trying to
find what that is.
In order to do that, we can
just use the integration method
and you'll see that we'll
get our final answer.
So, I can rewrite this as the integral.
So I can say Y equals to X
squared plus 3x minus one.
So I'm just expressing my intention
here by putting this integral sign.
So I'm showing that I know what I'm doing.
And my method that I'm going
to use is to integrate.
And I'll put a DX at the end.
And I don’t want you to worry about this
dX because it's not really doing anything.
It's just a statement.
It's saying, "with respect to X."
Remember the definition that
we've just gone over previously.
If you have X to the power of n dX,
you are adding 1 to the power and
you are dividing by the new power,
and then we obviously put a plus C at
the end for an indefinite integral.
So now, we should do that
with every single term.
Look at your x squared.
In order to integrate X squared,
we're going to add one to the power and
we're going to divide by the new power.
So this gives me X to the 3,
so 2 plus 1, and then
you divide it by 3.
And there we've integrated x squared.
For our next term, add 2 to the power.
So remember this is to the power of 1.
So this goes to 3x 2 over 2.
For our next term, we now
have a constant minus 1.
Remember what we did when we
differentiate it, it just disappeared.
But let me just write that here.
If I have one by itself, it's the
same as 1x to the power of 0.
If I'm trying to integrate this,
remember you add
1 to the power.
So, 0 plus 1 just gives you
X to the 1 divided by 1.
So this time, 1 doesn’t disappear.
One or any constant
actually integrates to x.
So minus 1 will integrate to minus x.
And then don’t forget that when we're
dealing with indefinite integrals,
we get a plus C at the end.
So, we're saying that our Y function
must have been x cube over 3,
plus 3x squared over 2,
minus x plus C.
Now, in this question, particularly,
we can't find our C yet.
Because in order to find C, we need
some values, we need some conditions.
We need to know what X we have been when Y
was a certain value to be able to find it.
But let's not let that bother us.
We'll just leave it like this.
This is our function and C is any
constant, so C is any number.
I can prove to you that this is
correct by differentiating this.
So what we could do is
differentiate this function now
because we're good
And when we differentiate,
if we've done it correctly,
we should get this
And that hopefully should convince you
that integration is the exact opposite
of the differentiation method.
So let's just try it out although that's --
we've answered our question here.
We're just going to give that a go.
So if we now dY by dX this,
remember what we said
when we differentiate,
bring the power down,
decrease the power by one.
So that goes to 3x squared and
the 3 is still at the bottom.
Bring the power down.
Decrease the power by 1.
So that gives me 6x to
the power of 1 over 2.
The two stays there.
This is x to the one.
So if you bring it down, we
just get 1x to the zero.
And then the C or any
constant just goes to zero.
If you tidy this up,
we can cancel the 3s.
We can cancel this with this to give you 3.
We end up with x squared
plus 3x to the 1 minus 1.
And look at this closely, we saw this two
minutes ago, that's this function here.
So it really is quite incredible
how closely linked integration
and differentiation is.
They're both parts of calculus
but they go from one to another.
So if you've differentiated something,
you can go back to your original
function by integrating,
you've integrated, you can differentiate
to prove just like we have here.