calculate-0-1-3-x-2-1-x-3-dx-

Question Number 50416 by Abdo msup. last updated on 16/Dec/18

c a l c u l a t e \int_{0}^{1}^{3} \sqrt{x^{2} (1 - x^{3})} d x

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

\int_{0}^{1} {x^{2} (1 - x^{3})}^{\frac{1}{3}} d x

t = x^{3} d t = 3 x^{2} d x

\frac{d t}{3 {(t)}^{\frac{2}{3}}} = d x

\int_{0}^{1} \frac{{t^{\frac{2}{3}} (1 - t)}^{\frac{1}{3}} \times \frac{d t}{3 t^{\frac{2}{3}}}}{1}

\frac{1}{3} \int_{0}^{1} t^{\frac{2}{9} - \frac{2}{3}} {(1 - t)}^{\frac{1}{3}} d t

\frac{1}{3} \int_{0}^{1} t^{\frac{2 - 6}{9}} {(1 - t)}^{\frac{1}{3}} d t

w a i t n o w u s e b e t a f u n c t i o n \dots

\frac{1}{3} \int t^{\frac{5}{9} - 1} {(1 - t)}^{\frac{4}{3} - 1} d t

f o r m u l a \int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} d x = β (m, n)

= \frac{⌈ (m) ⌈ (n)}{⌈ (m + n)}

a n s w e r i s = \frac{1}{3} \times \frac{⌈ (\frac{5}{9}) ⌈ (\frac{4}{3})}{⌈ (\frac{5}{9} + \frac{4}{3})}

= \frac{1}{3} \times \frac{⌈ (\frac{5}{9}) ⌈ (\frac{4}{3})}{⌈ (\frac{17}{9})}

w a i t b u s y \dots