Convergence or Divergence proving: $\int_a^b x\sqrt\frac{x-a}{b-x}dx$

Question

I have some integral, b > a $$\int_a^b x\sqrt\frac{x-a}{b-x}dx$$ I tried to use the comparison test, but can't find right integral.

Which way I can prove Convergence or Divergence this integral?

Answer 1

There exist a constant $A > 0$ such that \[ x\sqrt{x - a} \leq A \] for all $x \in [a, b]$.

Thus, the given integral is bounded by \[ \leq A \int_{a}^{b}(b - x)^{-1/2}dx. \]

Answer 2

HINT

Substitution $$\sqrt\frac{x-a}{b-x} = t,\quad x = {a+bt^2\over 1+t^2} = b+{a-b\over 1+t^2},\quad dx={2(b-a)\over(1+t^2)^2}\,tdt$$ gives converges integral $$\int\limits_a^b x\sqrt\frac{x-a}{b-x}dx = \int\limits_0^\infty {2(a+bt^2)t^2\over(1+t^2)^3}dt$$