∫_( 0) ^(2a) ((f(x))/(f(x)+f(2a−x))) dx =


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Question Number 42410 by soufiane zarik last updated on 25/Aug/18

$\underset{0}{\int^{2 a}} \frac{f (x)}{f (x) + f (2 a - x)} d x =$
Commented by maxmathsup by imad last updated on 25/Aug/18

$l e t A = \int_{0}^{2 a} \frac{f (x)}{f (x) + f (2 a - x)} d x c h a n g e m e n t 2 a - x = t g i v e$ $A = \int_{0}^{2 a} \frac{f (2 a - t)}{f (2 a - t) + f (t)} d t \Rightarrow 2 A = \int_{0}^{2 a} \frac{f (x)}{f (x) + f (2 a - x)} d x + \int_{0}^{2 π} \frac{f (2 a - x)}{f (2 a - x) + f (x)} d x$ $= \int_{0}^{2 a} \frac{f (x) + f (2 a - x)}{f (2 a - x) + f (x)} d x = \int_{0}^{2 a} d x = 2 a \Rightarrow A = a .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18

	$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x f o r m u l a$ $I = \int_{0}^{2 a} \frac{f (x)}{f (x) + f (2 a - x)} d x$ $= \int_{0}^{2 a} \frac{f (2 a - x)}{f (2 a - x) + f (x)} d x$ $a d d i n g 2 I = \int_{0}^{2 a} 1 . d x$ $2 I = 2 a$ $I = a$
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