Soln check:Int evaluation w/ sub rule

**Foxlion** · April 26th 2012, 03:45 PM

$\int_1^2 x\sqrt{x-1}dx$

I used the substituttion $u^2 = x-1 \Rightarrow 2udu = dx \Rightarrow x = u^2 + 1 \Rightarrow u(2) = 1; u(1) = 0$

and the integral then becomes $\int_0^1u(u^2+1)2udu \Rightarrow 2\int_0^1u^4 +u^2$ correct??

**Foxlion** · April 26th 2012, 03:57 PM

I'm sorry for the random updates pressed 'enter ' TOO EARLY by mistake

**Soroban** · April 26th 2012, 04:10 PM

Hello, Foxlion!

Lookin' good!

**skeeter** · April 26th 2012, 04:23 PM

or ...

$\int_1^2 x\sqrt{x-1} \, dx$

$u = x-1$

$x = u+1$

$du = dx$

$\int_0^1 (u+1)\sqrt{u} \, du$

$\int_0^1 u^{3/2} + u^{1/2} \, du$

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