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

1 Answer
Alexander
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#

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