00:01
We have another definite integral to evaluate here with the fundamental theorem of calculus.
00:04
I'm going to start off by integrating this, so increasing the power and dividing.
00:08
So 2t to the power 4 over 4 minus 6 t cubed over 3.
00:14
We'll keep our limits on here, 3 to 0.
00:16
It's all unusual that the lower limit's bigger than the top limit, but we can work with it.
00:21
Okay, we'll simplify this down first.
00:23
This is t to the power 4 over 2, minus 2, t cubed.
00:29
Okay, we'll substitute our two limits in, starting with the top limit, which was the zero...