∫_0 ^2 (√((4−x)/x))−(√(x/(4−x))) dx ?


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Question Number 90784 by jagoll last updated on 26/Apr/20

$\underset{0}{\int^{2}} \sqrt{\frac{4 - x}{x}} - \sqrt{\frac{x}{4 - x}} d x ?$
Commented by jagoll last updated on 26/Apr/20

$\int_{0}^{2} \frac{{(\sqrt{4 - x})}^{2} - {(\sqrt{x})}^{2}}{\sqrt{x} (\sqrt{4 - x})} =$ $\int_{0}^{2} \frac{4 - 2 x}{\sqrt{4 x - x^{2}}} d x = \int_{0}^{2} \frac{d (4 x - x^{2})}{\sqrt{4 x - x^{2}}}$ ${= 2 \sqrt{4 x - x^{2}}]}_{0}^{2} = 2 \sqrt{8 - 4}$ $= 2 \times 2 = 4$
Commented by abdomathmax last updated on 27/Apr/20
$A=∫_0 ^2 (√((4−x)/x))dx−∫_0 ^2 (√(x/(4−x)))dx =H−K changement (√((4−x)/x))=t give ((4−x)/x)=t^2 ⇒ 4−x =t^2 x ⇒(1+t^2 )x =4 ⇒x=(4/(t^2 +1)) ⇒ H =−∫_1 ^(+∞) t×(−((4(2t))/((t^2 +1)^2 )))dt =8 ∫_1 ^(+∞) ((t^2 +1−1)/((t^2 +1)^2 ))dt =8 ∫_1 ^(+∞) (dt/(1+t^2 ))−8∫_1 ^(+∞) (dt/((1+t^2 )^2 )) ∫_1 ^(+∞) (dt/(1+t^2 )) =(π/2)−(π/4) =(π/4) ∫_1 ^(+∞) (dt/((1+t^2 )^2 )) =_(t=tanθ) ∫_(π/4) ^(π/2) ((1+tan^2 θ)/((1+tan^2 θ)^2 ))dθ =∫_(π/4) ^(π/2) cos^2 θ =(1/2)∫_(π/4) ^(π/2) (1+cos(2θ))dθ =(1/2)×(π/4) +(1/4)[sin(2θ)]_(π/4) ^(π/2) =(π/8)+(1/4)(−1) ⇒ H =2π−8{(π/8)−(1/4)}=π+2 and K=∫_0 ^2 (√(x/(4−x)))dx we do the changement (√(x/(4−x)))=t ⇒ (x/(4−x)) =t^2 ⇒x =4t^2 −t^2 x ⇒(1+t^2 )x =4t^2 ⇒ x =((4t^2 )/(t^2 +1)) ⇒(dx/dt) =((8t(t^2 +1)−4t^2 (2t))/((t^2 +1)^2 ))=((8t)/((t^2 +1)^2 )) K =∫_0 ^1 t×((8t)/((t^2 +1)^2 ))dt =8 ∫_0 ^1 ((t^2 +1−1)/((t^2 +1)^2 ))dt =8 ∫_0 ^1 (dt/(t^2 +1))−8 ∫_0 ^1 (dt/((t^2 +1)^2 )) =8×(π/4)−8{ ∫_0 ^(π/4) (((1+tan^2 θ)dθ)/((1+tan^2 θ)^2 ))} (t=tanθ) =2π −8{∫_0 ^(π/4) cos^2 θ dθ} =2π−4∫_0 ^(π/4) (1+cos(2θ))dθ =π−2 [sin(2θ)]_0 ^(π/4) =π−2{1} =π−2 A =H−K =π +2−π +2 =4$
$A = \int_{0}^{2} \sqrt{\frac{4 - x}{x}} d x - \int_{0}^{2} \sqrt{\frac{x}{4 - x}} d x = H - K$ $c h a n g e m e n t \sqrt{\frac{4 - x}{x}} = t g i v e \frac{4 - x}{x} = t^{2} \Rightarrow$ $4 - x = t^{2} x \Rightarrow (1 + t^{2}) x = 4 \Rightarrow x = \frac{4}{t^{2} + 1} \Rightarrow$ $H = - \int_{1}^{+ \infty} t \times (- \frac{4 (2 t)}{{(t^{2} + 1)}^{2}}) d t$ $= 8 \int_{1}^{+ \infty} \frac{t^{2} + 1 - 1}{{(t^{2} + 1)}^{2}} d t = 8 \int_{1}^{+ \infty} \frac{d t}{1 + t^{2}} - 8 \int_{1}^{+ \infty} \frac{d t}{{(1 + t^{2})}^{2}}$ $\int_{1}^{+ \infty} \frac{d t}{1 + t^{2}} = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}$ $\int_{1}^{+ \infty} \frac{d t}{{(1 + t^{2})}^{2}} =_{t = t a n θ} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{1 + {t a n}^{2} θ}{{(1 + {t a n}^{2} θ)}^{2}} d θ$ $= \int_{\frac{π}{4}}^{\frac{π}{2}} {c o s}^{2} θ = \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} (1 + c o s (2 θ)) d θ$ $= \frac{1}{2} \times \frac{π}{4} + \frac{1}{4} {[s i n (2 θ)]}_{\frac{π}{4}}^{\frac{π}{2}} = \frac{π}{8} + \frac{1}{4} (- 1) \Rightarrow$ $H = 2 π - 8 {\frac{π}{8} - \frac{1}{4}} = π + 2 a n d K = \int_{0}^{2} \sqrt{\frac{x}{4 - x}} d x$ $w e d o t h e c h a n g e m e n t \sqrt{\frac{x}{4 - x}} = t \Rightarrow$ $\frac{x}{4 - x} = t^{2} \Rightarrow x = 4 t^{2} - t^{2} x \Rightarrow (1 + t^{2}) x = 4 t^{2} \Rightarrow$ $x = \frac{4 t^{2}}{t^{2} + 1} \Rightarrow \frac{d x}{d t} = \frac{8 t (t^{2} + 1) - 4 t^{2} (2 t)}{{(t^{2} + 1)}^{2}} = \frac{8 t}{{(t^{2} + 1)}^{2}}$ $K = \int_{0}^{1} t \times \frac{8 t}{{(t^{2} + 1)}^{2}} d t = 8 \int_{0}^{1} \frac{t^{2} + 1 - 1}{{(t^{2} + 1)}^{2}} d t$ $= 8 \int_{0}^{1} \frac{d t}{t^{2} + 1} - 8 \int_{0}^{1} \frac{d t}{{(t^{2} + 1)}^{2}}$ $= 8 \times \frac{π}{4} - 8 {\int_{0}^{\frac{π}{4}} \frac{(1 + {t a n}^{2} θ) d θ}{{(1 + {t a n}^{2} θ)}^{2}}} (t = t a n θ)$ $= 2 π - 8 {\int_{0}^{\frac{π}{4}} {c o s}^{2} θ d θ}$ $= 2 π - 4 \int_{0}^{\frac{π}{4}} (1 + c o s (2 θ)) d θ$ $= π - 2 {[s i n (2 θ)]}_{0}^{\frac{π}{4}} = π - 2 {1} = π - 2$ $A = H - K = π + 2 - π + 2 = 4$
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