Can someone tell me how to do these?
Estimate INT from 0 to 1 2/(1+x^2)dx by subdividing the interval into 8 parts, using
(i) the left Riemann sum: INT from 0 to 1 2/(1+x^2)dx
(ii) the right Riemann sum: INT from 0 to 1 2/(1+x^2)dx
(iii) the trapezoid rule INT from 0 to 1 2/(1+x^2)dx
I'm not surprised this took a while to get an answer. It involves a lot of calculation, and I personally like to help most with a problem that I can knock off in a minute or two. That said, I'll have a crack at it. (Thank you, perl!)
Divide the interval into 8 parts. While not necessary, let's use equal intervals of 1/8 = 0.125 So, the values for x are
0, 0.125, 0.250, 0.375, 0.050, 0.625, 0.750, 0.875, 1.000
The left sum will be construct rectangles with the left side at each of the first 8 x-values, then use f(x) as the height. The sum will be the total of all those areas.
Below is a table, listing the x value, the interval width, f(x), that box's area, and the accumulated area up to that point.
0.000 0.125 2.000 0.250 0.250
0.125 0.125 1.969 0.246 0.496
0.250 0.125 1.882 0.235 0.731
0.375 0.125 1.753 0.219 0.951
0.500 0.125 1.600 0.200 1.151
0.625 0.125 1.438 0.180 1.330
0.750 0.125 1.280 0.160 1.490
0.875 0.125 1.133 0.142 1.632
Now, for the right sum, we use the right side of each rectangle as the height.
0.125 0.125 1.969 0.246 0.246
0.250 0.125 1.882 0.235 0.481
0.375 0.125 1.753 0.219 0.701
0.500 0.125 1.600 0.200 0.901
0.625 0.125 1.438 0.180 1.080
0.750 0.125 1.280 0.160 1.240
0.875 0.125 1.133 0.142 1.382
1.000 0.125 1.000 0.125 1.507
Now, the trapezoid rule Averages the left and right heights. Shown here is the average x-value, average height for each trapezoid.
0.062 0.125 1.985 0.248 0.248
0.188 0.125 1.926 0.241 0.489
0.312 0.125 1.818 0.227 0.716
0.438 0.125 1.677 0.210 0.926
0.562 0.125 1.519 0.190 1.116
0.688 0.125 1.359 0.170 1.285
0.812 0.125 1.206 0.151 1.436
0.938 0.125 1.066 0.133 1.569
The exact value is π/2 = 1.571
The trapezoid rule nicely splits the left ad right Riemann sums