### Video Transcript

Evaluate the double integral shown
here.

Now when we’re looking to evaluate
this double integral, what it is is actually an integral inside another
integral. And I can rewrite it like this. So, we’ve got the definite integral
of one. And that’s with the limits two and
negative one, inside the definite integral with limits one and negative one. But this time we’ll be taking a
look at why.

So, the first thing we’ll do is
we’re gonna integrate. And we’ve got a one here. If there was a 𝑦 term involved,
then the 𝑦 term would just be a constant value. And we wouldn’t deal with it with
this integral. However, as we’ve just got one,
we’re gonna integrate this and see what the result is.

So, when we’ve integrated our one,
we get 𝑥. But then how do we work out a
definite integral? Well, to remind ourselves what to
do, what we do if we want the definite integral of 𝑓 of 𝑥 between the limits 𝑏
and 𝑎 is we integrate. And then we substitute in 𝑏
instead of 𝑥. And then we subtract from this the
integral with 𝑎 substituted in instead of 𝑥.

So, when we do that, inside the
parentheses, we’re gonna get two minus negative one. And that’s because we’ve
substituted in our bounds for 𝑥. So, now we’re left with an
integral. And this is another definite
integral. And this time, it’s between the
limits of one and negative one. And this time, it’s the integral of
three. And we’re looking at d𝑦 here. So, we’re looking in terms of
𝑦.

So, now what we’ve done is we’ve
completed the second integral of our double integral. And when we integrate three, we get
three 𝑦. Cause remember, this time we’re
dealing with 𝑦. So, we’ve now got three 𝑦. And what we need to do is we
substitute in one for 𝑦 and negative one for 𝑦. So, when we do this, we’re gonna
get three multiplied by one minus three multiplied by negative one. So therefore, this is gonna give us
an answer of six.

So, we can say that when we have
evaluated the double integral, as shown, the value is going to be six. And we achieved that by dealing
with it as an integral inside an integral. So, we first integrated and we were
dealing in 𝑥. Then we next integrated dealing in
𝑦, each time using the bounds shown.

And the question here is, why would
we use a double integral? Well, the double integral gives us
the volume below the surface of 𝑓 of 𝑥 comma 𝑦 above a region R. And the region R will be a
rectangular region that we’d look out. So, that’s what we get if you found
the double integral.