2-0-4x-x-3-1-3-dx-

Question Number 127521 by mohammad17 last updated on 30/Dec/20

{\int_{0}}^{2} \sqrt[3]{4 x - x^{3}} d x

Answered by Dwaipayan Shikari last updated on 30/Dec/20

\int_{0}^{2} \sqrt[3]{x} (\sqrt[3]{4 - x^{2}}) d x x^{2} = 4 u \Rightarrow x = 2 \frac{d u}{d x}

= 2 \int_{0}^{1} x^{- \frac{2}{3}} {(4 - 4 u)}^{\frac{1}{3}} d u = 2 \int_{0}^{1} u^{- \frac{1}{3}} {(1 - u)}^{\frac{1}{3}} d u = 2 β (\frac{2}{3}, \frac{4}{3})

= 2 \frac{Γ (\frac{2}{3}) Γ (\frac{1}{3})}{3 . Γ (2)} = \frac{2 π}{3 s i n (\frac{π}{3})} = \frac{4 π}{3 \sqrt{3}}

Answered by mathmax by abdo last updated on 01/Jan/21

I = \int_{0}^{2}^{3} \sqrt{4 x - x^{3}} dx \Rightarrow I = - \int_{0}^{2} x (^{3} \sqrt{- \frac{4}{x^{2}} + 1}) dx

changement \frac{2}{x} = \frac{1}{t} \Rightarrow x = 2 t \Rightarrow I = - \int_{0}^{1} (2 t) {(- \frac{1}{t^{2}} + 1)}^{\frac{1}{3}} (2 dt)

= - 4 \int_{0}^{1} t {(\frac{- 1 + t^{2}}{t^{2}})}^{\frac{1}{3}} dt = - 4 \int_{0}^{1} \frac{t}{t^{\frac{2}{3}}} {(t^{2} - 1)}^{\frac{1}{3}} dt

= 4 \int_{0}^{1} t^{\frac{1}{3}} {(1 - t^{2})}^{\frac{1}{3}} dt =_{t = \sqrt{u}} 4 \int_{0}^{1} u^{\frac{1}{6}} {(1 - u)}^{\frac{1}{3}} \frac{du}{2 \sqrt{u}}

= 2 \int_{0}^{1} u^{\frac{1}{6} - \frac{1}{2}} {(1 - u)}^{\frac{1}{3}} du = 2 \int_{0}^{1} u^{- \frac{1}{3}} {(1 - u)}^{\frac{1}{3}} du = 2 B (\frac{2}{3}, \frac{4}{3})

= 2 \frac{Γ (\frac{2}{3}) Γ (\frac{4}{3})}{Γ (\frac{2}{3} + \frac{4}{3})} = \frac{2}{Γ (2)} \times Γ (\frac{2}{3}) Γ (\frac{4}{3})

Commented by mathmax by abdo last updated on 01/Jan/21

I = 2 Γ (\frac{2}{3}) . Γ (\frac{1}{3} + 1) = 2 Γ (\frac{2}{3}) \times \frac{1}{3} Γ (\frac{1}{3})

= \frac{2}{3} Γ (\frac{1}{3}) . Γ (1 - \frac{1}{3}) = \frac{2}{3} \times \frac{π}{\sin (\frac{π}{3})} = \frac{2 π}{3 \times \frac{\sqrt{3}}{2}} = \frac{4 π}{3 \sqrt{3}}