Hi, everyone.
Welcome to integralcalc.com.
Today we’re going to be talking doing a double integrals problem.
And the one we’re going to be dealing with is the integral from 0 to 3 of the integral
from 0 to 3 again of xy + 7x + y dx dy.
The first thing that we should look at is the order at which our dx dy appears.
Sometimes we have dx and then dy.
Sometimes we have dy and then dx.
What we care about is recognizing that dx is going to match this inside integral over
here from zero to 3.
And then we’ve got an outer integral from 0 to 3 that relates to dy.
And basically, what these brackets tells us is that we always want to work from the inside
out.
So first, because dx is on the inside, we’re going to be taking the integral with respext
to x first from 0 to 3 based on this inner integral limits of integration.
And then once we’re done with that and we’ve plugged in our limits of integration and simplified
as much as we can, then we can take the integral with respect to y from 0 to 3 based on the
outer integral.
Inorder to take the integral of xy + 7x + y with respect to x, we’ll be treating y
as a constant.
In this case here we can go ahead and take the outer integral and leave that alone for
now.
Then I’m going to drop a big bracket and take the integral of everything inside here.
The integral of xy.
First of all, we recognize that y wilkl be treated like a constant on the x variable.
This first term xy, when we’re taking the integral of that with respect to x and holding
y constant, it’s basically the same thing as saying that we’re taking the integral
of something like 4x.
With y as a constant, it becomes a coefficient on the x term just like 4x.
And if we’re going to take the integral of 4x, we would get 4/2 x^2.
Same thing here.
We’re going to divide by 2 on the coefficient and then we’re going to square the x term
and we’re just going to leave the y because the 4 stays where it is.
It’s the same thing so the y is going to stay right where it is.
So, 1/2 (x^2)y.
The integral of 7x with respect to x is easy because there’s no y involved so that’s
just going to be 7/2 x^2.
And then the integral with respect to y.
Imagine if you’re taking a regular integral with only x variables and you’ve got the
integral of 4.
The integral of that is just 4x.
You just tack on that variable.
Because this is a constant then you add the variable.
That’s the integral.
So similarly here, y is just like our constant 4 so we’re just going to add the x term
to it and the integral will be xy. [1/2 (x^2)y + 7/2 x^2 + xy]
So that’s our integral with respect to x but we have to evaluate from 0 to 3 based
on these inner limits of integration and then we’re still going to have to integrate with
respect to y later.
For now, let’s go ahead and simplify what we just did with the integral with respect
to x.
We’ll plug in our limits of integration.
The thing to note is, because we just took the integral with respect to x, we’re going
to be plugging in for x and leaving y as it is.
So, we’ll plug in 3 for x and leave y right where it is.
So plus 7/2 (3^2) + 3y and then we subtract from that what we get when we plug in 0.
So 1/2 (0^2)y + 7/2 (0^2) + 0y and then dy here.
If we simplify this, that will give us 9/2 y + 63/2 + 3y.
And then these terms will be zero because everything’s multiplied by zero and then
we go ahead and write our dy.
This is essentially as simple as we can get, this integral here that we took with respect
to x.
Now that we’ve simplified as much as we can, we can go ahead and take the integral
with respect to y.
Taking the integral with respect to y, we’ll get y^2 and then we’d get 9/2 divided by
2 which will be 9/4 y^2 and then 63/2, we’ll just add the y variable to that because the
integral of a constant is just that constant plus that variable.
Then the integral of 3y, we’ll get 3/2 y^2 and now we want to evaluate that on the range
zero to 3.
That’s our integral with respect to y and now it’s just a matter of plugging in our
limits of integration for y and we don’t have any x variables left but if we did, we
would do the same thing and we would just plug in for y and leave x as it was.
Let’s actually o ahead and combine these y terms here.
If we multiply this last term by 2/2, we would get 6/4 + 9/4 = 15/4 y^2 + 63/2 y
on the range 0 to 3.
Now we can very easily plug in our limits of integration.
So 15/4 (3^2) + 63/2 (3) and then we subtract from that what we get when we plug in zero.
So 15/4 (0^2) + 63/2 (0) and that’s it.
let’s simplify.
We get 135/4 + 189/2 and these will go to zero so let’s go ahead and multiply this
second term by 2/2 so that we could get a common denominator.
We’ll get 135/4 + 378/4 = 513/4.
That’s it.
I don’t think you could simplify 513/4 but again, we took the integral with respect to
x, plugged in our inner limits of integration, 0 to 3, simplified tat as much as we could
and then took the integral with respect to y, evaluated again on those outer limits of
integration and we’re able to simplify to our final answer of 513/4.
I hope that video helped you guys and I will see you on the next one.
Bye!