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Question Number 43825 by Necxx last updated on 15/Sep/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Sep/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 17/Sep/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 17/Sep/18

	$c a l c u l a t i n g t h e v a l u e o f d . . . r e q u i r e d a n s i s$ $a + b + c = a + b + c + d - d . . . r e f e r t h e d i a g r a m$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Sep/18
	$d=∫_0 ^2 (x/2)dx+∫_2 ^5 5±(√(25−(x−5)^2 )) dx =∫_0 ^2 (x/2)dx+∫_2 ^5 5±(√(25−(x−5)^2 )) dx =∣(x^2 /4)∣_0 ^2 +5∣x∣_2 ^5 ±I =1+15±I =16±I ∫_2 ^5 (√(25−(x−5)^2 )) dx ∣((x−5)/2)(√(25−(x−5)^2 )) +((25)/2)sin^(−1) (((x−5)/5))∣_2 ^5 upper limit gives value 0 putting lower limit {((−3)/2)×4+((25)/2)sin^(−1) (−(3/5))} =−6+12.5sin^(−1) (((−3)/5)) =−6+12.5(−((37×Π)/5)) =−6−2.57Π I=0−{−6−2.57Π} I=6+2.57Π 16±(6+2.57Π) =22+2.57Π and 10−2.57Π value of d 22+2.57Π ←not feasible so d=10−2.57Π so required answer is a+b+c+d−d =21.50−(10−2.57𝚷) =11.50+2.57Π pls check...$
	$d = \int_{0}^{2} \frac{x}{2} d x + \int_{2}^{5} 5 \pm \sqrt{25 - {(x - 5)}^{2}} d x$ $= \int_{0}^{2} \frac{x}{2} d x + \int_{2}^{5} 5 \pm \sqrt{25 - {(x - 5)}^{2}} d x$ $=∣ \frac{x^{2}}{4} ∣_{0}^{2} + 5 ∣ x ∣_{2}^{5} \pm I$ $= 1 + 15 \pm I$ $= 16 \pm I$ $\int_{2}^{5} \sqrt{25 - {(x - 5)}^{2}} d x$ $∣ \frac{x - 5}{2} \sqrt{25 - {(x - 5)}^{2}} + \frac{25}{2} {s i n}^{- 1} (\frac{x - 5}{5}) ∣_{2}^{5}$ $u p p e r l i m i t g i v e s v a l u e 0$ $p u t t i n g l o w e r l i m i t$ ${\frac{- 3}{2} \times 4 + \frac{25}{2} {s i n}^{- 1} (- \frac{3}{5})}$ $= - 6 + 12.5 {s i n}^{- 1} (\frac{- 3}{5})$ $= - 6 + 12.5 (- \frac{37 \times Π}{5})$ $= - 6 - 2.57 Π$ $I = 0 - {- 6 - 2.57 Π}$ $I = 6 + 2.57 Π$ $16 \pm (6 + 2.57 Π)$ $= 22 + 2.57 Π a n d 10 - 2.57 Π$ $v a l u e o f d 22 + 2.57 Π \leftarrow n o t f e a s i b l e$ $s o d = 10 - 2.57 Π$ $s o r e q u i r e d a n s w e r i s$ $a + b + c + d - d$ $= 21.50 - (10 - 2.57 Π)$ $= 11.50 + 2.57 Π$ $p l s c h e c k . . .$
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