Welcome to the first example of
changing the order of integration of a double integral.
Taking a look at this integral here,
notice we're going to be in respect to y first,
which means that y is greater than or equal to x,
and less than or equal to one.
Let's go ahead and write that out.
We know y is greater than or equal to
the lower limit of integration, x,
and it's also less than or equal to
the upper limit of integration of one,
and also that x is on the closed integral from zero to one
where that x is greater than or equal to zero,
and it's less than or equal to one.
Let's go ahead and graph this region of integration.
So first, let's graph y is greater than or equal to x.
That's a line through the origin with a slope of one,
so it'd pass through this point here
and this point here.
And notice, it states that y is greater than or equal to x,
so we would shade above this line,
and then, y is also less than or equal to one.
Well, here's the line, y equals one,
and we would shade below the line,
and this point of intersection here
would be the point one, one.
And then, x is greater than or equal to zero,
that'd be along here,
and to the right,
and it's also less than or equal to one,
so from here to the left.
So, they can see, this would be our region of integration
for this double integral.
If we take a close look at this integral,
we would not be able to integrate this in its current order.
Hopefully, when we switch the order of integration,
it'll be much more manageable.
Let's go ahead and give it a try.
Let's go ahead and label these lines again.
This was x equals zero,
this was y equals x,
and this was y equals one.
So now, if you want to integrate with respects to x first,
remember, our limits of integration
must be functions of y,
so if we integrate from left to right first,
the lower limit of integration would be zero.
The upper limit would be this line here,
which we're going to express as x equals y.
And then, for the limits of integration for the y,
we're going to integrate from zero, up to one.
So now, we integrate this with respects to x.
Y is treated as a constant, so this ends up just being
x squared over two, times the quantity,
one plus y to the third, to the 1/2.
And because we integrated with respects to x,
we're going to replace x with y and zero.
Let's go ahead and factor this 1/2 out,
and then, when x is equal to y, we'll have y squared,
times the quantity, one plus y to the third, dy.
Let's go ahead and finish this on the next slide.
Looks like we'll have to perform U substitution here,
this would be our u,
so differential u would be three y squared dy.
Looking at our integral, we do have a y squared dy,
so let's go ahead and divide this by three,
so y squared dy is the same as 1/3 du,
so in terms of u, we'd have 1/2 times 1/3.
There's our du,
and this just become u to the 1/2.
Looks like we'd have 1/6.
This would be u to the 3/2
divided by 3/2 or times 2/3,
and of course, u is one plus y to the third.
So, it looks like this simplifies nicely.
We have 1/9,
and then, when y is one, we have two to the 3/2,
minus, and then, when y is zero, one to the 3/2.
Simplifying this, we have a denominator of nine.
Two to the 3/2 is the same as two square-root two,
and this would be minus one.
So you can see, once we switch the order of integration,
we were able to integrate this
using our integration techniques.
I hope you found this example helpful.
Thank you for watching.