let f_n (a) =∫_0 ^a x^n (√(a^2 −x^2 ))dx with a>0 1) determine a explicit form of f(a) 2) let g_n (a) =f^′ (a) give g_n (a) at form of integral and give its value 3) find the value of ∫_0 ^2 x^3 (√(4−x^2 ))dx and ∫


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Question Number 61229 by maxmathsup by imad last updated on 30/May/19

$l e t f_{n} (a) = \int_{0}^{a} x^{n} \sqrt{a^{2} - x^{2}} d x w i t h a > 0$ $1) d e t e r m i n e a e x p l i c i t f o r m o f f (a)$ $2) l e t g_{n} (a) = f^{^{'}} (a) g i v e g_{n} (a) a t f o r m o f i n t e g r a l a n d g i v e i t s$ $v a l u e$ $3) f i n d t h e v a l u e o f \int_{0}^{2} x^{3} \sqrt{4 - x^{2}} d x a n d \int_{0}^{\sqrt{3}} x^{4} \sqrt{3 - x^{2}} d x$
	Answered by perlman last updated on 30/May/19

	$1) p u t x = a s i n (t)$ $f n (a) = a^{n + 1} \int_{0}^{\frac{π}{2}} {s i n}^{n} (t) \sqrt{(a^{2} - a^{2} {s i n}^{2} (t)} c o s (t) d t =$ $= a^{n + 2} \int_{0}^{\frac{π}{2}} {s i n}^{n} (t) {c o s}^{2} (t) d t = a \int {s i n}^{n} (t) d t - a^{n + 2} \int {s i n}^{n + 2} (t) d t$ $l e t I_{n} = \int_{0}^{\frac{π}{2}} {s i n}^{n} (t) d t = \int s i n (t) {s i n}^{n - 1} (t) d t = [- c o s (t) {s i n}^{n - 1} (t)] + (n - 1) \int {c o s}^{2} (t) {s i n}^{n - 2} (t) d t$ $(n - 1) \int_{0}^{\frac{π}{2}} (1 - {s i n}^{2} (t)) {s i n}^{(n - 2)} (t) d t = (n - 1) I_{n - 2} - (n - 1) I_{n} = I_{n}$ $I_{n} = \frac{n - 1}{n} I_{n - 2}$ $I_{0} = \frac{π}{2}$ $I_{1} = 1$ $I_{2 n} = \frac{2 n - 1}{2 n} I_{2 (n - 1)}$ $I_{2 n} = \frac{2 n - 1}{2 n} . \frac{2 (n - 1) - 1}{2 (n - 1)} . . . . . . \frac{2 - 1}{2} I_{0} = \frac{(2 n - 1) (2 n - 3) . . . . (1)}{2 n . 2 (n - 1) . . . . 2 (1)} I_{0}$ $= \frac{2 n (2 n - 1) (2 n - 2) . . . . . . . 1}{{[2^{n} n!]}^{2}} I_{0} = \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} \frac{π}{2}$ $I_{2 n + 1} = \frac{2 n}{2 n + 1} I_{2 n - 1} = \frac{2 n}{2 n + 1} . \frac{2 n - 2}{2 n - 1} . . . . . . \frac{2}{3} I_{1} = \frac{{(2^{n} n!)}^{2} . 2}{(2 n + 1)!} = \frac{2^{2 n + 1} {(n!)}^{2}}{(2 n + 1)!}$ $f n (a) = a^{n + 2} (I_{n} - I_{n + 2})$ $g n (a) = \frac{d}{d a} \int_{0}^{a} x^{n} \sqrt{(a^{2} - x^{2})} d x = \frac{d}{d a} \int_{0}^{1} a^{n} t^{n} a^{2} \sqrt{(1 - t^{2})} d t = \int_{0}^{1} \frac{d}{d a} (a^{n + 2} t^{n} \sqrt{(1 - t^{2}} d t) =$ $(n + 2) \int_{0}^{1} a^{n + 1} t^{n} \sqrt{(1 - t^{2}} d t$ $\int_{0}^{2} x^{3} \sqrt{(4 - x^{2})} d x$ $n = 3 a = 2$ $= 2^{5} (I 1 - I 3) = 2^{5} (1 - \frac{2}{3}) = \frac{32}{3}$
	Commented by maxmathsup by imad last updated on 31/May/19

	$t h a n k s s i r .$
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