find ∫_(−5) ^5 ((√(25−x^2 )))dx whithout using trigonometric compensation


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Question Number 106329 by 175mohamed last updated on 04/Aug/20

$f i n d \int_{- 5}^{5} (\sqrt{25 - x^{2}}) d x w h i t h o u t u s i n g$ $t r i g o n o m e t r i c c o m p e n s a t i o n$
	Answered by Her_Majesty last updated on 04/Aug/20

	$y = \sqrt{25 - x^{2}} i s a h a l f c i r c l e w i t h c e n t e r (0 / 0)$ $a n d r a d i u s 5 \Rightarrow a n s w e r i s \frac{25 π}{2}$
	Answered by Dwaipayan Shikari last updated on 04/Aug/20

	$\int_{- 5}^{5} \sqrt{25 - x^{2}} = 2 \int_{0}^{5} \sqrt{25 - x^{2}} = 2 {[\frac{x}{2} \sqrt{25 - x^{2}} + \frac{25}{2} {s i n}^{- 1} \frac{x}{5}]}_{0}^{5}$ $= \frac{25 π}{2}$
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