00:01
And we need to determine another definite integral.
00:02
So let's start by making this a bit easier for ourselves.
00:05
Since there's only one term in the place of the denominator, so yes, the denominator, we can rewrite this as 6 over t to the power 3, which is the same as 6t to the power negative 3.
00:23
Let's just put a bracket there.
00:25
And then minus t over t to the power 3 is the same as t to the power negative 2.
00:30
Dt.
00:31
Now that should be much easier to work with than trying to do some sort of weird trying to remember rules and whatnot, which probably won't even work anyway.
00:43
So let's go and do that then.
00:47
So this gets us well t to the power negative two so divide by a negative two is negative three t and then this gets us t to the power negative one divide by negative 1, so that's plus from 1 to 2.
01:08
And let's go substitute those values in.
01:11
So negative 3 times 2 to the power negative 2 plus 2 to the power negative 1, minus negative 3 times 1 to the power negative 2...