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$$\int\limits_{1}^{2} x\sqrt{x-1}\;\text{d}x.$$

I've set $u=x-1$ and rearranged the equation to get $x=u+1$. I differentiated $u=x-1$ to $\text{d}u=\text{d}x$.

I've rewritten the integral as $\int_{0}^{1} (u+1)\sqrt{u}\;\text{d}u$ which gives me 2 while the answer is $\frac{16}{15}$. I think I differentiated the substitute equation incorrectly. Could someone explain my mistake to me?

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  • $\begingroup$ Oh wow, I just forgot to take the antiderivative of u. Sorry for wasting everybody's time xD. $\endgroup$
    – Greener
    May 11, 2014 at 14:56
  • $\begingroup$ before doing anything, I would express the first $x$ as $(x-1)+1$. Shoulda had a V8 $\endgroup$
    – John Joy
    May 11, 2014 at 21:23

2 Answers 2

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$$\int_0^1(u+1)\sqrt u\ du=\int_0^1(u^{\frac32}+u^{\frac12})du$$

$$=\left(\frac{u^{\frac32+1}}{\frac32+1}+\frac{u^{\frac12+1}}{\frac12+1}\right)\big|_0^1$$

$$=\frac1{\frac52}+\frac1{\frac32}=\frac25+\frac23=\cdots$$

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Do $x - 1 = u^2$. If $x = 1$, then $u = 0$. If $x = 2$, then $u = 1$. Otherside, $dx = 2udu$.

$\int_{1}^{2}x\sqrt{x - 1}dx = \int_{0}^{1}(1 + u^2)u\cdot 2udu = 2\int_{0}^{1}(u^2 + u^4)du = \frac{16}{15}$

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