calculate I=∫_0 ^1 (√(1+2(√(x−x^2 ))))dx and J =∫


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Question Number 47985 by maxmathsup by imad last updated on 17/Nov/18

$c a l c u l a t e I = \int_{0}^{1} \sqrt{1 + 2 \sqrt{x - x^{2}}} d x a n d J = \int_{0}^{1} \sqrt{1 - 2 \sqrt{x - x^{2}}} d x$
Commented by MJS last updated on 18/Nov/18

$I = \frac{4}{3}$ $J = \frac{4}{3} - \frac{2 \sqrt{2}}{3}$ $I will post my work later$
Commented by maxmathsup by imad last updated on 18/Nov/18

$a l s o w e h a v e J = \int_{0}^{1} ∣ \sqrt{x} - \sqrt{1 - x} ∣ d x b u t x - (1 - x) = 2 x - 1 s o$ $i f 2 x - 1 ⩾ 0 \Rightarrow \frac{1}{2} ⩽ x ⩽ 1 w e g e t ∣ \sqrt{x} - \sqrt{1 - x} ∣= \sqrt{x} - \sqrt{1 - x} \Rightarrow$ $∣ \sqrt{x} - \sqrt{1 - x} ∣= \sqrt{x} - \sqrt{1 - x}$ $i f 2 x - 1 ⩽ 0 \Rightarrow 0 ⩽ x ⩽ \frac{1}{2} \Rightarrow∣ \sqrt{x} - \sqrt{1 - x} ∣= \sqrt{1 - x} - \sqrt{x} \Rightarrow$ $J = \int_{0}^{\frac{1}{2}} (\sqrt{1 - x} - \sqrt{x}) d x + \int_{\frac{1}{2}}^{1} (\sqrt{x} - \sqrt{1 - x}) d x$ $\int_{0}^{\frac{1}{2}} (\sqrt{1 - x} - \sqrt{x}) d x = {[- \frac{2}{3} {(1 - x)}^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}}]}_{0}^{\frac{1}{2}} = \frac{2}{3} - \frac{2}{3} {(\frac{1}{2})}^{\frac{3}{2}} - \frac{2}{3} {(\frac{1}{2})}^{\frac{3}{2}}$ $= \frac{2}{3} - \frac{4}{3} \frac{1}{2^{\frac{3}{2}}} = \frac{2}{3} - \frac{2^{2 - \frac{3}{2}}}{3} = \frac{2}{3} - \frac{\sqrt{2}}{3}$ $\int_{\frac{1}{2}}^{1} (\sqrt{x} - \sqrt{1 - x}) d x = {[\frac{2}{3} x^{\frac{3}{2}} + \frac{2}{3} {(1 - x)}^{\frac{3}{2}}]}_{\frac{1}{2}}^{1} = \frac{2}{3} - \frac{2}{3} {(\frac{1}{2})}^{\frac{3}{2}} - \frac{2}{3} {(\frac{1}{2})}^{\frac{3}{2}}$ $= \frac{2}{3} - \frac{4}{3} {(\frac{1}{2})}^{\frac{3}{2}} = \frac{2}{3} - \frac{\sqrt{2}}{3} \Rightarrow J = 2 (\frac{2}{3} - \frac{\sqrt{2}}{3}) = \frac{4}{3} - \frac{2 \sqrt{2}}{3} .$
Commented by maxmathsup by imad last updated on 18/Nov/18

$w e h a v e f i r s t 1 + 2 \sqrt{x - x^{2}} = 1 + 2 \sqrt{x} \sqrt{1 - x} = x + 1 - x + 2 \sqrt{x} \sqrt{1 - x}$ $= {(\sqrt{x} + \sqrt{1 - x})}^{2} \Rightarrow \sqrt{1 + 2 \sqrt{x - x^{2}}} =∣ \sqrt{x} + \sqrt{1 - x} ∣ b u t x \in [0, 1] \Rightarrow∣ \sqrt{x + \sqrt{1 - x}} ∣=$ $\sqrt{x} + \sqrt{1 - x} \Rightarrow I = \int_{0}^{1} (\sqrt{x} + \sqrt{1 - x}) d x = \int_{0}^{1} \sqrt{x} d x + \int_{0}^{1} \sqrt{1 - x} d x$ $= {[\frac{2}{3} x^{\frac{3}{2}}]}_{0}^{1} + {[- \frac{2}{3} {(1 - x)}^{\frac{3}{2}}]}_{0}^{1} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Nov/18
	$1+2(√(x−x^2 )) 1+2(√(x(1−x))) x+1−x+2(√(x(1−x))) {(√x) +(√(1−x)) }^2 I=∫_0 ^1 (√x) +(√(1−x)) dx now =∣(x^(3/2) /(3/2))∣_0 ^1 −(2/3)∣(1−x)^(3/2) ∣_0 ^1 =(2/3)+(2/3)=(4/3) J= ∫_0 ^(0.5) (√(1−2(√(x(1−x))))) dx+∫_(0.5) ^1 (√(1−2(√(x(1−x))))) dx ∫_0 ^(0.5) (√(1−x)) −(√x) dx+∫_(0.5) ^1 (√x) −(√(1−x)) dx from0.5 >x>0 value of (√(1−x)) >value of (√x) and from 1>x>0.5 valud of (√x) >(√(1−x)) so =∣{−(2/3)(1−x)^(3/2) −(2/3)(x)^(3/2) }∣_0 ^(0.5) +∣{(2/3)(x)^(3/2) +(2/3)(1−x)^(3/2) }∣_(0.5) ^1 =(−(2/3)){(0.5)^(3/2) +(0.5)^(3/2) −(1)^(3/2) −0}+(2/3){1+0−(0.5)^(3/2) −(0.5)^(3/2) } =(((−4)/3))(0.5)^(3/2) +(2/3)+(2/3)−((4/3))(0.5)^(3/2) =(4/3)−(8/3)×(0.5)^(3/2) =(4/3)−(8/3)×(1/(2(√2)))=(4/3)−(4/3)×(1/((√2) ))$
	$1 + 2 \sqrt{x - x^{2}}$ $1 + 2 \sqrt{x (1 - x)}$ $x + 1 - x + 2 \sqrt{x (1 - x)}$ ${\sqrt{x} + \sqrt{1 - x}}^{2}$ $I = \int_{0}^{1} \sqrt{x} + \sqrt{1 - x} d x$ $n o w$ $=∣ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} ∣_{0}^{1} - \frac{2}{3} ∣ {(1 - x)}^{\frac{3}{2}} ∣_{0}^{1} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$ $J =$ $\int_{0}^{0.5} \sqrt{1 - 2 \sqrt{x (1 - x)}} d x + \int_{0.5}^{1} \sqrt{1 - 2 \sqrt{x (1 - x)}} d x$ $\int_{0}^{0.5} \sqrt{1 - x} - \sqrt{x} d x + \int_{0.5}^{1} \sqrt{x} - \sqrt{1 - x} d x$ $f r o m 0.5 > x > 0 v a l u e o f \sqrt{1 - x} > v a l u e o f \sqrt{x}$ $a n d f r o m 1 > x > 0.5 v a l u d o f \sqrt{x} > \sqrt{1 - x}$ $s o$ $=∣ {- \frac{2}{3} {(1 - x)}^{\frac{3}{2}} - \frac{2}{3} {(x)}^{\frac{3}{2}}} ∣_{0}^{0.5} + ∣ {\frac{2}{3} {(x)}^{\frac{3}{2}} + \frac{2}{3} {(1 - x)}^{\frac{3}{2}}} ∣_{0.5}^{1}$ $= (- \frac{2}{3}) {{(0.5)}^{\frac{3}{2}} + {(0.5)}^{\frac{3}{2}} - {(1)}^{\frac{3}{2}} - 0} + \frac{2}{3} {1 + 0 - {(0.5)}^{\frac{3}{2}} - {(0.5)}^{\frac{3}{2}}}$ $= (\frac{- 4}{3}) {(0.5)}^{\frac{3}{2}} + \frac{2}{3} + \frac{2}{3} - (\frac{4}{3}) {(0.5)}^{\frac{3}{2}}$ $= \frac{4}{3} - \frac{8}{3} \times {(0.5)}^{\frac{3}{2}} = \frac{4}{3} - \frac{8}{3} \times \frac{1}{2 \sqrt{2}} = \frac{4}{3} - \frac{4}{3} \times \frac{1}{\sqrt{2}}$
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