How do you evaluate the integral int 1/sqrtxdx from 0 to 1?

1 Answer
Jul 11, 2016

int_0^1 (1)/(sqrt(x)) dx =[2sqrt(x)]_0^1 = (2sqrt(1)-2sqrt(0)) = 2

Explanation:

We can rewrite int_0^1 (1)/(sqrt(x)) dx as int_0^1 (x^(-1/2)) dx, which is now easier to see and evaluate.

int_0^1 (x^(-1/2)) dx = [(x^(-1/2 + 1/1)) * (1/ (-1/2+1/1))]_0^1

Since (-1/2) + (1/1) = 1/2, we have

int_0^1 (1)/(sqrt(x)) dx =[2sqrt(x)]_0^1 = (2sqrt(1)-2sqrt(0)) = 2