let-I-n-0-1-x-n-1-x-dx-1-calculate-I-0-and-I-1-2-prove-that-n-N-3-2n-I-n-2n-I-n-1-3-find-I-n-interms-of-n-

Question Number 40139 by maxmathsup by imad last updated on 16/Jul/18

l e t I_{n} = \int_{0}^{1} x^{n} \sqrt{1 - x} d x

1) c a l c u l a t e I_{0} a n d I_{1}

2) p r o v e t h a t \forall n \in N^{★} (3 + 2 n) I_{n} = 2 n I_{n - 1}

3) f i n d I_{n} i n t e r m s o f n

Commented by prof Abdo imad last updated on 20/Jul/18

1) I_{0} = \int_{0}^{1} \sqrt{1 - x} d x =_{x = {s i n}^{2} t} \int_{0}^{\frac{π}{2}} c o s t 2 s i n t c o s t ? d t

= 2 \int_{0}^{\frac{π}{2}} s i n t {c o s}^{2} t d t = 2 {[- \frac{1}{3} {c o s}^{3} t]}_{0}^{\frac{π}{2}}

= - \frac{2}{3} {0 - 1} = \frac{2}{3}

I_{1} = \int_{0}^{1} x \sqrt{1 - x} d x =_{\sqrt{1 - x} = t} - \int_{0}^{1} (1 - t^{2}) t (- 2 t d t)

= 2 \int_{0}^{1} t^{2} (1 - t^{2}) d t = 2 \int_{0}^{1} (t^{2} - t^{4}) d t

= 2 {[\frac{t^{3}}{3} - \frac{t^{5}}{5}]}_{0}^{1} = 2 {\frac{1}{3} - \frac{1}{5}} = 2 \frac{2}{15} = \frac{4}{15}

Commented by maxmathsup by imad last updated on 24/Jul/18

2) c h a n g e m e n t \sqrt{1 - x} = t g i v e 1 - x = t^{2}

I_{n} = \int_{0}^{1} {(1 - t^{2})}^{n} t 2 t d t = \int_{0}^{1} 2 t^{2} {(1 - t^{2})}^{n} d t

=_{t = s i n α} \int_{0}^{\frac{π}{2}} 2 {s i n}^{2} α {c o s}^{2 n} α c o s α d α

= \int_{0}^{\frac{π}{2}} 2 {s i n}^{2} α {c o s}^{2 n} α d α b y p a r t s u^{^{'}} = s i n α {c o s}^{2 n} α a n d v = 2 s i n α

I_{n} = {[- \frac{1}{2 n + 1} {c o s}^{2 n + 1} α 2 s i n α]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} - \frac{1}{2 n + 1} {c o s}^{2 n + 1} α . 2 c o s α d α

= \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} {c o s}^{2 n + 2} α d α = \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} (1 - {s i n}^{2} α) {c o s}^{2 n} α d α

= \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} {c o s}^{2 n} α d α - \frac{1}{2 n + 1} \int_{0}^{\frac{π}{2}} 2 {s i n}^{2} α {c o s}^{2 n} α d α

= \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} (1 - {s i n}^{2} α) {c o s}^{2 (n - 1)} α d α - \frac{1}{2 n + 1} I_{n} \Rightarrow

(1 + \frac{1}{2 n + 1}) I_{n} = \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} {c o s}^{2 (n - 1)} α d α - \frac{1}{2 n + 1} I_{n - 1} \Rightarrow

\frac{2 n + 2}{2 n + 1} I_{n} = \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} {c o s}^{2 (n - 1)} α d α - \frac{1}{2 n + 1} I_{n - 1} \dots b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 24/Jul/18

3) w e h a v e I_{n} = \frac{2 n}{2 n + 3} I_{n - 1} w i t h n ⩾ 1 \Rightarrow

\prod_{k = 1}^{n} I_{k} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 3} \prod_{k = 1}^{n} I_{k - 1} \Rightarrow

I_{1} . I_{2} . I_{3} \dots I_{n} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 3} I_{0} . I_{1} . I_{2} \dots . I_{n - 1} \Rightarrow

I_{n} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 3} . I_{0} = \frac{2}{3} \prod_{k = 1}^{n} \frac{2 k}{2 k + 3}