calculate ∫_0 ^2 (√(x^3 (2−x)))dx


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Question Number 40147 by maxmathsup by imad last updated on 16/Jul/18

$c a l c u l a t e \int_{0}^{2} \sqrt{x^{3} (2 - x)} d x$
Commented by math khazana by abdo last updated on 17/Jul/18
$let I = ∫_0 ^2 (√(x^3 (2−x)))dx I =∫_0 ^2 x(√(x(2−x))) changement x=2sin^2 θ give I = ∫_0 ^(π/2) 2 sin^2 θ(√(2sin^2 θ(2−2sin^2 θ))) 4 sinθ cosθdθ =16∫_0 ^(π/2) sin^2 θ sinθ cosθ sinθ cosθ dθ =16 ∫_0 ^(π/2) sin^4 θ cos^2 θdθdθ by parts ∫_0 ^(π/2) cosθ( cosθ sin^4 θ)dθ =[(1/5) sin^5 θ cosθ]_0 ^(π/2) +∫_0 ^(π/2) (1/5) sin^5 θ sinθ dθ =(1/5) ∫_0 ^(π/2) (sin^2 θ)^3 dθ =(1/5) ∫_0 ^(π/2) {((1−cos(2θ))/2)}^3 dθ =(1/(40)) ∫_0 ^(π/2) {Σ_(k=0) ^3 C_3 ^k (−1)^k (cos(2θ))^k }dθ =(1/(40)) ∫_0 ^(π/2) {1 −3cos(2θ) +3cos^2 (2θ) −cos^3 (2θ)}dθ =(π/(80)) −(3/(40)) ∫_0 ^(π/2) cos(2θ)dθ +(3/(40)) ∫_0 ^(π/2) ((1+cos(4θ))/2)dθ −(1/(40)) ∫_0 ^(π/2) cos(2θ)((1+cos(4θ))/2) dθ =(π/(80)) +(3/(160)) −(1/(80)) ∫_0 ^(π/2) (1+cos(4θ))cos(2θ) ....be continued...$
$l e t I = \int_{0}^{2} \sqrt{x^{3} (2 - x)} d x$ $I = \int_{0}^{2} x \sqrt{x (2 - x)} c h a n g e m e n t x = 2 {s i n}^{2} θ g i v e$ $I = \int_{0}^{\frac{π}{2}} 2 {s i n}^{2} θ \sqrt{2 {s i n}^{2} θ (2 - 2 {s i n}^{2} θ)} 4 s i n θ c o s θ d θ$ $= 16 \int_{0}^{\frac{π}{2}} {s i n}^{2} θ s i n θ c o s θ s i n θ c o s θ d θ$ $= 16 \int_{0}^{\frac{π}{2}} {s i n}^{4} θ {c o s}^{2} θ d θ d θ b y p a r t s$ $\int_{0}^{\frac{π}{2}} c o s θ (c o s θ {s i n}^{4} θ) d θ$ $= {[\frac{1}{5} {s i n}^{5} θ c o s θ]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} \frac{1}{5} {s i n}^{5} θ s i n θ d θ$ $= \frac{1}{5} \int_{0}^{\frac{π}{2}} {({s i n}^{2} θ)}^{3} d θ$ $= \frac{1}{5} \int_{0}^{\frac{π}{2}} {\frac{1 - c o s (2 θ)}{2}}^{3} d θ$ $= \frac{1}{40} \int_{0}^{\frac{π}{2}} {\sum_{k = 0}^{3} C_{3}^{k} {(- 1)}^{k} {(c o s (2 θ))}^{k}} d θ$ $= \frac{1}{40} \int_{0}^{\frac{π}{2}} {1 - 3 c o s (2 θ) + 3 {c o s}^{2} (2 θ) - {c o s}^{3} (2 θ)} d θ$ $= \frac{π}{80} - \frac{3}{40} \int_{0}^{\frac{π}{2}} c o s (2 θ) d θ + \frac{3}{40} \int_{0}^{\frac{π}{2}} \frac{1 + c o s (4 θ)}{2} d θ$ $- \frac{1}{40} \int_{0}^{\frac{π}{2}} c o s (2 θ) \frac{1 + c o s (4 θ)}{2} d θ$ $= \frac{π}{80} + \frac{3}{160} - \frac{1}{80} \int_{0}^{\frac{π}{2}} (1 + c o s (4 θ)) c o s (2 θ)$ $. . . . b e c o n t i n u e d . . .$
Commented by math khazana by abdo last updated on 25/Jul/18
$∫_0 ^(π/2) {1+cos(4θ)}cos(2θ)dθ =∫_0 ^(π/2) cos(2θ)dθ +∫_0 ^(π/2) cos(2θ)cos(4θ)dθ =0 +(1/2) ∫_0 ^(π/2) {cos(6θ) +cos(2θ)}dθ =0 ⇒ I =16{ ((3π)/(160))} ⇒ I =((3π)/(10)) .$
$\int_{0}^{\frac{π}{2}} {1 + c o s (4 θ)} c o s (2 θ) d θ$ $= \int_{0}^{\frac{π}{2}} c o s (2 θ) d θ + \int_{0}^{\frac{π}{2}} c o s (2 θ) c o s (4 θ) d θ$ $= 0 + \frac{1}{2} \int_{0}^{\frac{π}{2}} {c o s (6 θ) + c o s (2 θ)} d θ = 0 \Rightarrow$ $I = 16 {\frac{3 π}{160}} \Rightarrow I = \frac{3 π}{10} .$
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