00:01
In this question, in part a, we're going to evaluate this integral.
00:03
The integral from 0 to 2 of x squared times the quantity of 3x cubed plus 1 being raised to the 1 third power dx.
00:10
Here we're going to use the technique of u substitution.
00:15
Namely, i'm going to let u equal the inside of my composite function, 3x cubed plus 1.
00:24
Now, once i have a u, i figure out my du.
00:28
In this case, 9x squared dx.
00:31
Now, i don't see 9x squared dx anywhere, but i do have x squared dx.
00:36
So i can divide by 9.
00:39
I can say du over 9 is equal to x squared dx.
00:46
Now, at this point, i need to change my limits of integration.
00:50
Whenever i do a u substitution within a definite integral, i have to change my limits of integration.
00:57
So in the world of x, x went from 0 to 2.
01:02
Now, how about u? if x is 0, u would be 1.
01:08
If x was 2, what's my u? 2 cubed is 8.
01:12
8 times 3, 24, plus 1, 25.
01:17
So that we have an integral from 1 to 25.
01:22
Now, the 3x cubed plus 1 quantity to the 1 third power, that's u to the 1 third.
01:30
My x squared dx, that is du over 9.
01:34
1 ninth, constant multiple, comes along for the ride.
01:39
My antiderivative of u to the 1 third power is u to the 4 thirds over 4 thirds, which we evaluate 1 to 25.
01:51
Now, division by 4 thirds, multiplication by 3 quarters.
01:55
1 ninth times 3 quarters is 3 over 36, or 1 twelfth.
02:01
U to the 4 thirds, evaluate 1 to 25.
02:07
This is 1 twelfth times 25 to the 4 thirds minus 1 to the 4 thirds...