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Evaluate using the Fundamental Theorem of Calculus: $$\int_{7}^{10} \frac{xdx}{\sqrt{x-6}}$$

I am stuck with the x on top. I substitute $u = x-6$ and then $du = dx$. soo... $$\int_{1}^{4} (u)^{-1/2}xdu$$

is it okay to still have that x there or do I need to substitute something else? something maybe like... $x(x-6)^{-1/2}?$

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    $x=u-1$: there should be no $x$ left. Also, how did you calculate the new bounds for your integral? They are incorrect.2012-07-25
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    I apologize, I had the wrong denominator. I fixed it though!2012-07-25
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    To answer your question: no, it's not okay to have $x$ in there. When you make a change of variable, **all** instances of the old variable must be replaced in terms of the new variable somehow. The new integral should involve only the new variable, and not the old one.2012-07-25

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Substitute that $x$ with $x = u + 6,$ you get: $$\int \frac{u + 6}{\sqrt{u}} du = \int \sqrt{u} du + \int \frac{6}{\sqrt{u}} du.$$