How do you use Riemann sums to evaluate the area under the curve of f(x) = 2-x^2 on the closed interval [0,2], with n=4 rectangles using midpoint?

1 Answer
Jul 12, 2016

The Reimann sum with n=4 will evaluate to 1.375.

Explanation:

Since we are using the midpoint method to evaluate the area of 4 rectangles on the interval [0,2], we know that our rectangles will have side lengths 2/4 = 0.5 and f(x).

Our rectangles will have midpoints located at;
x = 0.25
x = 0.75
x = 1.25
x = 1.75

Thus our approximation of the area under f(x) = 2 -x^2 will be:
A = 0.5(f(0.25) + f(0.75) + f(1.25) + f(1.75))
A = 0.5(1.9375 + 1.4375 + 0.4375 - 1.0625)
A = 0.5(2.75) = 1.375

We can compare this to the result of an actual integral by evaluating:
int_0^2 2 - x^2 = [2x - 1/3 x^3]_0^2
= (4 - 8/3) - 0 = 4/3 = 1.33...

So as we can see, the difference is quite small, even for a small number of rectangles like n =4.