Video Transcript
Evaluate the double integral of
𝑥𝑦 between the limits two and zero and four and one with respect to 𝑥 and with
respect to 𝑦.
So when we’re looking to evaluate
this double integral, what it means is an integral inside another integral. So if we rewrite our integral, what
we’re gonna have is the definite integral of 𝑥𝑦 between the limits four and one
and this is with respect to 𝑥. Then this is inside the definite
integral between the limits of two and zero with respect to 𝑦. So the first thing we’re gonna do
is integrate. So we’re gonna integrate 𝑥𝑦. But because we did this with
respect to 𝑥, the 𝑦 just becomes a constant term. So it isn’t dealt with when we’re
dealing with the integration at this stage.
So when we integrate, we integrate
our 𝑥𝑦. And like I said, the 𝑦 just stays
constant. We’re gonna get 𝑥 squared over
two. And that’s because we’ve raised the
exponent of the 𝑥 term by one. So it becomes 𝑥 squared. And we divide by the new exponent,
which is two. So let’s just remind ourselves now
what we do to work out the definite integral. Well, if we want to work out the
definite integral of 𝑓 of 𝑥 between the limits 𝑏 and 𝑎, then what we do is we
integrate the 𝑓 of 𝑥, which we’re at that stage now. And then, we substitute in for 𝑥
the value 𝑏. And then, we subtract from this the
value of our integral with 𝑎 substituted in instead of 𝑥.
So when we do that, we get four
squared over two cause we’ve substituted in our four cause that’s the upper bound
multiplied by 𝑦 minus one squared over two multiplied by 𝑦. Well, for each of these terms,
we’re gonna get eight 𝑦 for the first term. And that’s because four squared is
16. 16 over two is eight, so eight
𝑦. Then, for the second term, we’re
gonna get half 𝑦. And that’s because one squared is
just one. So it will give us a half 𝑦. So now, what we have is the
definite integral of 15 over two 𝑦 between the limits of two and zero. And that’s because if you have
eight 𝑦 and you take away a half, that’s seven and a half 𝑦 or 15 over two 𝑦.
So now we just deal with this, but
this time, with respect to 𝑦. And it’s gonna be equal to 15 over
four 𝑦 squared with limits two and zero. And we get that cause if you
integrate 15 over two 𝑦. First of all, we raise the exponent
of 𝑦 by one. So we get 𝑦 squared. And then, we divide by the new
exponent. So we divide by two. Well, if we divide 15 over two by
two, that means it’s the same as multiplying the denominator by two, which gives us
four. So we get 15 over four 𝑦
squared. So now what we do is we substitute
in our upper and lower bounds. So we get 15 over four multiplied
by two squared minus 15 over four multiplied by zero squared. And this leads to 15 over four
multiplied by four. And that’s because if we subtract
15 over four, multiply it by zero squared. Well, this is just zero.
So what can we do now? Well, the fours cancel. And that’s because we were
multiplying by four and dividing by four. So you can remove those. So therefore, we can say that if we
evaluate our double integral, then the result is gonna be equal to 15. And as I said, we’ve now found the
value of the double integral. But why would we use a double
integral? Well, a double integral gives us
the volume below the surface of 𝑓 of 𝑥 comma 𝑦 above a region 𝑅. And that region 𝑅 is a rectangular
region. So that’s why we would use a double
integral.
