# evaluating $\int_{7}^{10} \frac{xdx}{\sqrt{x-6}}$

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. –  Olivier Bégassat Jul 25 '12 at 4:56
I apologize, I had the wrong denominator. I fixed it though! –  user69 Jul 25 '12 at 4:58
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. –  Arturo Magidin Jul 25 '12 at 5:01

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