
Nitin P. answered 05/27/20
Machine Learning Engineer - UC Berkeley CS+Math Grad
Consider the function f(x) = x-1/2 . The improper integral over [0,1] evaluates to:
∫01 x-1/2dx = [2√x]01 = 2
However, we have f2(x) = 1/x, which results in:
∫01 x-1dx = [ln x]01 = 0 - ln 0 = ∞
Therefore, f is integrable but f2 is not integrable.

Nitin P.
All we're doing here is evaluating two integrals and showing that one is finite whereas the other is not. In this case, we see that 1/sqrt(x) has a finite improper integral, because its antiderivative is defined at 0. However, when we integrate 1/x, we get ln x, which is not defined at x = 0. Therefore, the former is integrable over the interval whereas the latter is not.05/28/20
Fatima A.
Thank you for caring but i feel it is kinda ambiguous for me, can you please explain the steps05/28/20