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Question Number 50007 by Joel578 last updated on 13/Dec/18

$For a < x < b, find$ $\underset{a}{\int^{b}} \sqrt{x - a} . \sqrt{b - x} d x$
Commented byAbdo msup. last updated on 13/Dec/18

$c h a n g e m e n t \sqrt{x - a} = t g i v e x - a = t^{2} \Rightarrow$ $I = \int_{0}^{\sqrt{b - a}} t \sqrt{b - (a + t^{2})} 2 t d t = 2 \int_{0}^{\sqrt{b - a}} t^{2} \sqrt{b - a - t^{2}} d t$ $=_{t = \sqrt{b - a} s i n u} 2 \int_{0}^{\frac{π}{2}} (b - a) {s i n}^{2} u \sqrt{b - a} c o s u \sqrt{b - a} c o s u d u$ $= 2 {(b - a)}^{2} \int_{0}^{\frac{π}{2}} {s i n}^{2} u {c o s}^{2} u d u$ $= 2 {(b - a)}^{2} \frac{1}{4} \int_{0}^{\frac{π}{2}} {s i n}^{2} (2 u) d u$ $= \frac{{(b - a)}^{2}}{2} \int_{0}^{\frac{π}{2}} \frac{1 - c o s (4 u)}{2} d u$ $= \frac{{(b - a)}^{2}}{4} \int_{0}^{\frac{π}{2}} (1 - c o s (4 u)) d u$ $= \frac{π}{8} {(b - a)}^{2} - \frac{{(b - a)}^{2}}{16} {[s i n (4 u)]}_{0}^{\frac{π}{2}}$ $★ I = \frac{π}{8} {(b - a)}^{2} ★$
	Answered by ajfour last updated on 13/Dec/18

	$l e t x - \frac{a + b}{2} = t$ $\int_{- \frac{b - a}{2}}^{\frac{b - a}{2}} \sqrt{t + \frac{b - a}{2}} \sqrt{\frac{b - a}{2} - t} d t$ $l e t c = \frac{b - a}{2}$ $= \int_{- c}^{c} \sqrt{c^{2} - t^{2}} d t = 2 \int_{0}^{c} \sqrt{c^{2} - t^{2}} d t$ $= 2 {[\frac{t}{2} \sqrt{c^{2} - t^{2}} + \frac{c^{2}}{2} \sin^{- 1} \frac{t}{c}]}_{0}^{c}$ $= \frac{c^{2} π}{2} = \frac{{(b - a)}^{2} π}{8} .$
	Commented byMJS last updated on 13/Dec/18

	$grea solutiont!$
	Commented byajfour last updated on 13/Dec/18

	$W h e n w e c o m e a c r o s s m a t t e r s$ $t h a t d o n o t r e q u i r e o u r a t t e n t i o n$ $a t t h i s m o m e n t, a l t h o u g h a r e$ $n e c e s s a r y f o r o u r c o n c l u s i o n,$ $t h e n i t i s b e t t e r t o r e p r e s e n t t h e m$ $b y t h e b r i e f e s t p o s s i b l e s y m b o l s .$ $- R e n e D e s c a r t e s (D i s c o u r s e o n M e t h o d)$
	Commented byJoel578 last updated on 13/Dec/18

	$v e r y w e l l S i r$
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