00:01
Let's evaluate the integral from 0 to 2 over 3 of the square root of 4 minus 9x squared.
00:08
Now let's look at this integrand.
00:10
This would be a little easier to work with if it was of the form 4 minus x squared because this is 2 squared minus x squared and we would like to be in the case a squared minus x squared.
00:27
But we're not in this case because there's a 9 in the front of the x.
00:31
So what we can do is rewrite this 4 minus 9x squared as 4 minus 3x squared.
00:44
Then we can do a u substitution.
00:47
U equals 3x, du equals 3dx, so that du over 3.
01:00
And the limits of integration will change because we did a u sub.
01:04
So x equals 0 was the original lower limit that's going to correspond to u equals 3 times 0, which is 0, and the upper limit was x equals 2 over 3.
01:21
And that's going to correspond to the upper limit, u equals 3 times 2 over 3, which is 2.
01:29
So this integral can be written as one -third.
01:35
The third is coming from the u -sub.
01:40
Integral, we have our new limits, 0 to 2, square root, 4 minus 3x squared, which is u squared, du.
01:56
So now we have this new integral, and this, we have a 4 minus u squared.
02:03
So this suggests that we should use the substitution, u equals 2 -san theta.
02:17
Then we have du equals 2 cosine theta, d theta.
02:25
And we have two options, two ways to proceed here because this is a definite integral.
02:30
One way to proceed, which we'll use next, is to find the new limits of integration through the trick substitution.
02:42
If you go this route, you won't have to draw the right triangle because you'll never need to go back to the variable u after the trig sub.
02:53
So here, let's observe that when you make a trig sub of this form, that you're putting a restriction on data.
03:00
So this is from the textbook.
03:02
We have data is between negative pi over 2 and pi over 2.
03:10
So the lower limit before was x, or excuse me, u after we did the u sub, was u equal 0.
03:20
That's going to correspond to plugging that's into the trig sub equation.
03:24
We have 0 equals 2 sine theta.
03:31
So this means sign is 0.
03:35
And the only time sign is 0 in this interval, negative pi over 2 to pi over 2, is when data is 0.
03:42
So this will be our new lower limit of integration.
03:45
Now for the upper limit was 2.
03:51
So u equals 2...