Question

${\int }_{\alpha }^{\beta }\sqrt{\frac{x-\alpha }{\beta -x}}dx=$

• A. $\frac{{\pi }^2}{2}(\beta -\alpha )$
• B. ${\int }_{\alpha }\beta \sqrt{\frac{\beta -x}{x-\alpha }}dx$
• C. $\frac{\pi }{2}(\beta -\alpha )$
• D. ${\int }_{\alpha }\beta \sqrt{\frac{\beta +x}{x+\alpha }}dx$

Answer 1

Answer: (B), (C)

${\int }_{\alpha }\beta \sqrt{\frac{\beta -x}{x-\alpha }}dx$
$\frac{\pi }{2}(\beta -\alpha )$

Let $l={\int }_{\alpha }^{\beta }\sqrt{\frac{x-\alpha }{\beta -x}}dx$

Substitute $\beta -x=t^2\Rightarrow dx=-2tdt$

$\therefore I={\int }_{\alpha }^{\beta }\sqrt{\frac{x-\alpha }{\beta -x}}dx=2{\int }_0^{\sqrt{\beta -a}}\sqrt{\sqrt{(\beta -\alpha {)}^2}-t^{2}dt}$

$=2{[\frac{1}{2}\sqrt{(\beta -\alpha )-t^2}+\left(\frac{\beta -\alpha }{2}\right]si{n}^{-1}\left[\frac{1}{\sqrt{\beta -\alpha }}\right)]}_0^{\sqrt{\beta -\alpha }}$

$=2[0+\frac{\beta -\alpha }{2}si{n}^{-1}]=2\times \left(\frac{\beta -\alpha }{2}\right)\frac{\pi }{2}=\frac{\pi }{2}(\beta -\alpha )$

And by using ${\int }_{\alpha }^{\beta }f\left(x\right)dx={\int }_{\alpha }^{\beta }f(a+b-x)dx$

${\int }_{\alpha }^{\beta }\sqrt{\frac{x-\alpha }{\beta -x}}dx={\int }_{\alpha }^{\beta }\sqrt{\frac{\beta -x}{x-\alpha }}dx$