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

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Answer 1 · 2016-10-27T09:30:56

The integral =2

We use the integral #intx^n=x^(n+1)/(n+1)# for #n!=-1#

So werewrite the integral

#int_0^1dx/sqrtx=int_0^1x^(-1/2)dx=(x^(1/2)/(1/2))_0^1#

#=2-0=2#

Answer 2 · 2016-10-27T13:47:58

This is an improper integral. The integrand is not defined at one point of the closed interval #[0,1]#.

Because the integrand in not defined at #0#, we (try to) evaluate the definite integral by using a limit of definite integrals.

#int_0^1 1/sqrtx dx = lim_(ararr0) int_a^1 1/sqrtx dx # if the limit exists.

#int_a^1 1/sqrtx dx = {:2sqrtx ]_a^1 = 2-2sqrta#.

So
#int_0^1 1/sqrtx dx = lim_(ararr0) (2-2sqrta) dx #

# = 2-2sqrt0 = 2#

Answer 3 · 2016-10-27T14:14:49

The integrand #1/sqrt x# is continuous in #[0_+, 1].#.

It does not exist at x = 0.

So, the integral is 2, for the limits #0_+ and 1 only.

It is improper to state that the value is 2, for the limits 0 and 1.

Valid integration:

#int 1/sqrt x dx#, for x from #0_+ to 1#

#= 2[sqrtx], between the limits

#= 2(1-#the right limit 0) =2

This is a problem in limits.