00:02
So problem 24, we're dealing with the limit definition of an integral.
00:06
So a lot of algebra involved in solving these guys.
00:09
I find it easiest just to break these up, so it breaks some of the algebra up, into two separate integrals.
00:15
So 2x dx minus the integral from 0 to 2 of x cubed dx.
00:23
Let's go do the work of each of those separate, and then we'll come right back here to find our answer.
00:29
So let's deal with the first integral first.
00:31
The integral from 0 to 2, so the integral from 0 to 2 of 2x dx by definition is going to be 2 times the limit as n approaches infinity of the sum of i equals 1 to n.
00:53
Now the width of each rectangle is just going to be 2 minus 0 over n.
00:58
So the width of each rectangle is 2 over n.
01:01
The height of each rectangle is going to be this function x evaluated so x at 2 over n 4 over n and so forth you're going to increment this so this is going to be 2 i over n so this is the limit or it's going to be 2x2 or it's going to be 2 times the limit as n approaches infinity and if you look at what i have here this is going to be you can replace that i with the formula that you know the sum i equal one to n of i is just simply in n plus 1 over 2 so that's going to take the sum of this out the summation sign so i'm going to be left with 2 over n and then times 2 n n plus 1 over 2 n so all i did was replace the what you see here in n plus 1 over 2 that replaces the i so in this case if i look at what i have to simplify this guy this is 2 times the limit as n approaches infinity and so what you see here is that you've got 2n over 2n and then if you look at this so this is going to be 2 and if i look at this in over n plus 1, i can write that as 1 plus 1 over n.
02:50
And so as n goes to infinity, this term will go to 0.
02:54
So it just becomes 2 times 2 times 1 plus 0, which is simply 4.
03:07
So if i look at where i am in the scope of this problem, i have determined that the value of this one is 4...