00:01
Consider the integral from 3 to 6 of x times e raised to negative x over 2 dx.
00:05
For the first part, we want to find our points a sub 0 until a sub 3 and x sub 1 until x sub 3, where a subn are endpoints of the sub intervals, well, x subn are midpoints of the subinterval.
00:26
So to do that, we need to find delta x first, which is b minus a over n.
00:32
If we subdivide the interval 3 to 6 into 3 sub -intervals, then n is equal to 3, and we have 6 minus 3 over 3 equal to 1.
00:41
So a -sub -0 is going to be 3 for the first endpoint, and then one unit from that will be our a -sub 1, which is 4.
00:51
A -sub 2 is 5, and a -sub -3 is 6.
00:56
And then we want to find the midpoints.
00:59
So x sub 1 is going to be 3 plus 4 over 2, which is 3 .5.
01:08
And since the points are always a unit away, then x sub 2 is 4 .5 and x sub 3 is 5 .5.
01:21
So for the value of the function, f of a0 will be f of 3, and that's 3 e raised to negative 3 over 2.
01:34
F of a sub 1 is f of 4 or 4, e raised to negative 4 over 2 or negative 2.
01:44
F of a sub 2 is f of 5, and that's 5 e raised to negative 2.
01:51
Negative 5 over 2, and f of a sub 3 is f of 6, which is 6 times e raised to negative 6 over 2 or negative 3.
02:03
And then for the midpoints, we have f of 3 .5 equal to 3 .5 e raised to negative 3 .5 over 2, f of 4 .5 is 4 .5e raised negative 4 .5 over 2.
02:28
And f of 5 .5 is 5 .5 e raised 2, negative 5 .5 over 2.
02:36
Now, using trapezoidal rule, you have area equal to 1 half of delta x times f of a0 plus 2f of 0, plus 2f of, of a sub 1 plus 2f of a sub 2 plus f of a sub 3...