Question

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

Answer 1

`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`