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 ankit24 February 17th, 2018 09:58 PM

Calculus: Integration

${\displaystyle\int_0^1}\! \sqrt{x\ln\small\left(\frac{1}{\large x}\right)}dx$

 skipjack February 17th, 2018 10:57 PM

According to W|A, it's $\displaystyle \frac13\sqrt{\frac{2\pi}{3}}$.

 ankit24 February 18th, 2018 07:08 AM

elaborate solution

 ankit24 February 18th, 2018 08:21 AM

also, if square root is replaced by cube root what will be the solution?

 skipjack February 18th, 2018 09:42 AM

Have you tried using a substitution?

 romsek February 18th, 2018 09:43 AM

Quote:
 Originally Posted by skipjack (Post 588755) Have you tried using a substitution?
or anything at all?

 skipjack February 18th, 2018 06:30 PM

Using $x = e^{-u^2}$, where $u\geqslant0$, so that $dx = -2ue^{-u^2}du$, one gets

$\displaystyle \int_\infty^0\! -2u^2e^{-3u^2/2}du = -(2/3)\!\int_0^\infty\! -3u^2e^{-3u^2/2}du$.

Using integration by parts, with $v = e^{-(3/2)u^2}$, so that $dv = -3ue^{-(3/2)u^2}du$, one gets

$\displaystyle -(2/3)\left[ue^{-(3/2)u^2}\right]_0^\infty + (2/3)\!\int_0^\infty\! e^{-(3/2)u^2}du = (2/3)\!\int_0^\infty\! e^{-(3/2)u^2}du$.

If $u = \sqrt{\frac23}t$, so that $du = \sqrt{\frac23}dt$, one gets $\frac23\sqrt{\frac23}$ $\displaystyle \!\int_0^\infty\! e^{-t^2}dt$.

It's well known that $\displaystyle \!\int_0^\infty\! e^{-t^2}dt = \small\frac{\sqrt\pi}{2}$.

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