Using Riemann Sums to approximate the area underneath a curve?

You need to specify the number of sub-intervals. Let us assume 8 sub-intervals.

You can change it to any number.

There are two Riemann sums (left and right). You also need to specify left or right. I have computed them both.

deltax =0.25

number of intervals 8

The intervals are:

[a(0),a(1)][a(1),a(2)][a(2),a(3)][a(3),a(4)]

[a(4),a(5)][a(5),a(6)][a(6),a(7)][a(7),a(8)]

(0,0.25)(0.25,0.5)(0.5,0.75)(0.75,1)

(1,1.25)(1.25,1.5)(1.5,1.75)(1.75,2)

The Right Riemann sum =

f(0.25)+f(0.5)+f(0.75)+

f(1)+f(1.25)+f(1.5)+f(1.75)+

f(2)

[1.0625]+[1.25]+[1.5625]+

[2]+[2.5625]+[3.25]+[4.0625]+

[5]

sum 20.75

deltax =0.25

Multiply sum by deltax = 5.1875 ---- Right Riemann sum

The Left Riemann sum =

f(0)+f(0.25)+f(0.5)+f(0.75)+

f(1)+f(1.25)+f(1.5)+f(1.75)

[1]+[1.0625]+[1.25]+[1.5625]+

[2]+[2.5625]+[3.25]+[4.0625]

deltax =0.25

sum 16.75

Multiplying by deltax = 4.1875

The Midpoint Rule =

f(0.125)+f(0.375)+f(0.625)+f(0.875)+

f(1.125)+f(1.375)+f(1.625)+f(1.875)

[1.01563]+[1.14063]+[1.39063]+[1.76563]+

[2.26563]+[2.89063]+[3.64063]+[4.51563]

deltax =0.25

sum 18.625

Multiplying by deltax = 4.6563 ----- left Riemann sum

i think of you like the relationship between the indispensable as an andiderivative and since the area under a curve. think of you're on the curve of a few function f(x). you initiate at a and flow alongside. you're taking small steps to get from one element on the curve to a element dx to the terrific suited in the x-path. for each step, undergo in techniques how some distance you went in the y-path. ultimately you attain x=b and are status on f(b). What happens in case you upload up all the y-path displacements for each step? You get the gap between f(b) and f(a), i.e. f(b) - f(a). each y-path displacement may be approximated (o.k., for small displacements) by way of the by-manufactured from f at that element, elevated by way of dx. This quickly leverages the definition of the by-product because of the fact the slope of the function. So, f(b) - f(a) ~= sum f'(x_i) dx. Taking the decrease because of the fact the step length is going to 0 makes the terrific suited area precisely the indispensable and makes equality carry precisely instead of roughly. in case you replace f with it fairly is antiderivative, F, and in addition replace f' with f, then we get F(b) - F(a) ~= sum f(x_i) dx, which may be quickly interpreted because of the fact the area under the curve f, of which F is the antiderivative. Or, we would desire to have interpreted the above because of the fact the area under f'. it fairly is a satisfied accident, that the sums may be interpreted because of the fact the area of rectangles or as vertical displacements. i like to think of of the fundamental theorem of calculus as "the place you're is the sum of the place you have been going".