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.

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