Hi, everyone.
Welcome back to itnegralcalc.com.
Today, we’re going to be doing a double integrals problem.
This one is the integral from 0 to 2 of the integral from 0 to 4 of 3x + 4y dx dy.
With any double integrals problem, the first thing we nee to do is recognize that we’ve
got two sets of notation.
One is the inner integral and the other one is the outer integral.
So if we draw brackets that way, we can see that the first integral is going to be taken
with respect to x because we’ve got dx here inside the first integral.
We’re going to completely ignore the outside for now, the integral from 0 to 2 and we’re
going to ignore dy.
And we’re just going to look at what’s inside the orange bracket.
So we’re going to be taking the integral with respect to dx because dx is this inner
most derivative notation.
We need to take the integral of 3x + 4y with respect to x from 0 to 4.
Taking the derivative with respect to x, we will ignore the 0 to 2 for now so we’re
just going to write it out there but then drop a big bracket so that we can kind of
take the section of what we’re going to do.
So to take the integral with respect to x of 3x is not challenging.
It’s just like a regular integral.
We would add 1 to the exponent which is an implied 1 so 1 + 1 = 2, x^2.
And then we divide the coefficient 3 by the new exponent 2.
So 3/2 x^2 is the integral of 3x.
Taking the integral of 4y with respect to x is a little bit trickier but not too bad.
So remember that if we’re just taking a normal integral, the most basic integral you
can think of, and you’ve got 3x + 4, the integral would be 3/2 x ^2 + 4x because this
4 here is a constant and we would just tack on the x variable to the constant.
In this case, because we’re just taking the integral with respect to x, we’re treating
y as a constant.
So 4 time y is also going to be a constant, which means that to take the integral of that,
we’re just going to tack on an x variable just like we did here.
So the integral would be 4y times x which of course just simplifies to 4xy.
So that’s the integral and then we’re going to be evaluating that on the range 0
to 4 given the limits of integration of our inside integral here.
SO we’re going to do that and then we’ve just got dy written out here.
We don’t need the dx again because we just took the integral with respect to x.
So this dx notation goes away.
We still need to continue simplifying what’s inside of our integral before we go ahead
and take the integral with respect to y.
Doing so, we’ll plug in 4 first.
4^2 + 4, and we’re plugging it in for x because we took the integral with respect
to x and leaving y as it is.
So we plug in for x, leave y as it is and we subtract what we get when we plug in zero.
So 0^2 + 4(0)y, and then dy.
Now, 4^2 = 16.
16(3) = 48.
48/2 = 24 so this whole term here simplifies to 24 + 16y.
This is going to reduce to 0 and this is going to reduce to 0.
SO we’ve fully simplified what’s inside the integral.
Now that we’ve fully simplified this, we can proceed to taking the integral with respect
to y So doing that of course will give us 24y just like we’ve said up here.
When we’ve got the constant inside, we just add the variable.
Now, with respect to y, we’re just going to add the y variable to the constant 24.
And then as for 16y, we’re going to add 1 to the exponent. 1 + 1 = 2.
We divide the coefficient 16 by the new exponent 2 and our integral is 16/2 y^2.
And we’re going to be evaluating that on the range 0 to 2.
Okay, so now we’re going to go ahead and plug in our limits of integration.
This is actually going to simplify to 24y + 8y^2 on the range 0 to 2.
Alright.
So 24(2) + 8(2)^2 minus what we get when we plug in zero so 24(0) + 8(0)^2 and that’ll
give us 48 + 32.
This will reduce to zero and this will reduce to zero so we don’t have to worry about
that.
48 + 32 = 80.
So that’s it.
That’s our final answer.
This double integral simplifies to 80.
I hope that video helped you guys and I will see you in the next one.
Bye!