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


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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
$1) I_0 =∫_0 ^1 (√(1−x))dx =_(x=sin^2 t) ∫_0 ^(π/2) cost 2sint cost?dt =2 ∫_0 ^(π/2) sint cos^2 t dt =2[−(1/3) cos^3 t]_0 ^(π/2) =−(2/3){0−1}=(2/3) I_1 = ∫_0 ^1 x(√(1−x))dx =_((√(1−x))=t) −∫_0 ^1 (1−t^2 )t(−2tdt) =2 ∫_0 ^1 t^2 (1−t^2 )dt =2 ∫_0 ^1 (t^2 −t^4 )dt =2 [(t^3 /3) −(t^5 /5)]_0 ^1 =2{(1/3) −(1/5)}=2(2/(15)) =(4/(15))$
$1) I_{0} = \int_{0}^{1} \sqrt{1 - x} d x =_{x = {s i n}^{2} t} \int_{0}^{\frac{π}{2}} c o s t 2 s i n t c o s t ? d t$ $= 2 \int_{0}^{\frac{π}{2}} s i n t {c o s}^{2} t d t = 2 {[- \frac{1}{3} {c o s}^{3} t]}_{0}^{\frac{π}{2}}$ $= - \frac{2}{3} {0 - 1} = \frac{2}{3}$ $I_{1} = \int_{0}^{1} x \sqrt{1 - x} d x =_{\sqrt{1 - x} = t} - \int_{0}^{1} (1 - t^{2}) t (- 2 t d t)$ $= 2 \int_{0}^{1} t^{2} (1 - t^{2}) d t = 2 \int_{0}^{1} (t^{2} - t^{4}) d t$ $= 2 {[\frac{t^{3}}{3} - \frac{t^{5}}{5}]}_{0}^{1} = 2 {\frac{1}{3} - \frac{1}{5}} = 2 \frac{2}{15} = \frac{4}{15}$
Commented by maxmathsup by imad last updated on 24/Jul/18

$2) c h a n g e m e n t \sqrt{1 - x} = t g i v e 1 - x = t^{2}$ $I_{n} = \int_{0}^{1} {(1 - t^{2})}^{n} t 2 t d t = \int_{0}^{1} 2 t^{2} {(1 - t^{2})}^{n} d t$ $=_{t = s i n α} \int_{0}^{\frac{π}{2}} 2 {s i n}^{2} α {c o s}^{2 n} α c o s α d α$ $= \int_{0}^{\frac{π}{2}} 2 {s i n}^{2} α {c o s}^{2 n} α d α b y p a r t s u^{^{'}} = s i n α {c o s}^{2 n} α a n d v = 2 s i n α$ $I_{n} = {[- \frac{1}{2 n + 1} {c o s}^{2 n + 1} α 2 s i n α]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} - \frac{1}{2 n + 1} {c o s}^{2 n + 1} α . 2 c o s α d α$ $= \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} {c o s}^{2 n + 2} α d α = \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} (1 - {s i n}^{2} α) {c o s}^{2 n} α d α$ $= \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} {c o s}^{2 n} α d α - \frac{1}{2 n + 1} \int_{0}^{\frac{π}{2}} 2 {s i n}^{2} α {c o s}^{2 n} α d α$ $= \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} (1 - {s i n}^{2} α) {c o s}^{2 (n - 1)} α d α - \frac{1}{2 n + 1} I_{n} \Rightarrow$ $(1 + \frac{1}{2 n + 1}) I_{n} = \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} {c o s}^{2 (n - 1)} α d α - \frac{1}{2 n + 1} I_{n - 1} \Rightarrow$ $\frac{2 n + 2}{2 n + 1} I_{n} = \frac{2}{2 n + 1} \int_{0}^{\frac{π}{2}} {c o s}^{2 (n - 1)} α d α - \frac{1}{2 n + 1} I_{n - 1} . . . b e c o n t i n u e d . . .$
Commented by maxmathsup by imad last updated on 24/Jul/18

$3) w e h a v e I_{n} = \frac{2 n}{2 n + 3} I_{n - 1} w i t h n ⩾ 1 \Rightarrow$ $\prod_{k = 1}^{n} I_{k} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 3} \prod_{k = 1}^{n} I_{k - 1} \Rightarrow$ $I_{1} . I_{2} . I_{3} . . . I_{n} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 3} I_{0} . I_{1} . I_{2} . . . . I_{n - 1} \Rightarrow$ $I_{n} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 3} . I_{0} = \frac{2}{3} \prod_{k = 1}^{n} \frac{2 k}{2 k + 3}$
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