Question

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?

Answer 1

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`.