let-f-n-a-0-a-x-n-a-2-x-2-dx-with-a-gt-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-

FacebookTweetPin

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

$$let{f}_n\left(a\right)={\int }_0^a{x}^n\sqrt{a^2-x^2}dxwitha>0$$$$1)determineaexplicitformoff\left(a\right)$$$$2)let{g}_n\left(a\right)=f^{{}^{\prime }}\left(a\right)give{g}_n\left(a\right)atformofintegralandgiveits$$$$value$$$$3)findthevalueof{\int }_0^2{x}^3\sqrt{4-x^2}dxand{\int }_0^{\sqrt{3}}{x}^4\sqrt{3-x^2}dx$$

Answered by perlman last updated on 30/May/19

$$1)putx=asin\left(t\right)$$$$fn\left(a\right)=a^{n+1}{\int }_0^{\frac{\pi }{2}}{sin}^n\left(t\right)\sqrt{(a^2-a^2{sin}^2\left(t\right)}cos\left(t\right)dt=$$$$=a^{n+2}{\int }_0^{\frac{\pi }{2}}{sin}^n\left(t\right){cos}^2\left(t\right)dt=a\int {sin}^n\left(t\right)dt-a^{n+2}\int {sin}^{n+2}\left(t\right)dt$$$$let{I}_n={\int }_0^{\frac{\pi }{2}}{sin}^n\left(t\right)dt=\int sin\left(t\right){sin}^{n-1}\left(t\right)dt=[-cos\left(t\right){sin}^{n-1}\left(t\right)]+(n-1)\int {cos}^2\left(t\right){sin}^{n-2}\left(t\right)dt$$$$(n-1){\int }_0^{\frac{\pi }{2}}(1-{sin}^2\left(t\right)){sin}^{(n-2)}\left(t\right)dt=(n-1)I_{n-2}-(n-1)I_n=I_n$$$$I_n=\frac{n-1}{n}{I}_{n-2}$$$$I_0=\frac{\pi }{2}$$$$I_1=1$$$$I_{2n}=\frac{2n-1}{2n}{I}_{2(n-1)}$$$$I_{2n}=\frac{2n-1}{2n}.\frac{2(n-1)-1}{2(n-1)}\dots \dots \frac{2-1}{2}{I}_0=\frac{(2n-1)(2n-3)\dots .\left(1\right)}{2n.2(n-1)\dots .2\left(1\right)}{I}_0$$$$=\frac{2n(2n-1)(2n-2)\dots \dots .1}{{[2^{n}n!]}^2}{I}_0=\frac{\left(2n\right)!}{2^{2n}{(n!)}^2}\frac{\pi }{2}$$$$I_{2n+1}=\frac{2n}{2n+1}{I}_{2n-1}=\frac{2n}{2n+1}.\frac{2n-2}{2n-1}\dots \dots \frac{2}{3}{I}_1=\frac{{(2^{n}n!)}^2.2}{(2n+1)!}=\frac{2^{2n+1}{(n!)}^2}{(2n+1)!}$$$$fn\left(a\right)=a^{n+2}(I_n-I_{n+2})$$$$gn\left(a\right)=\frac{d}{da}{\int }_0^a{x}^n\sqrt{(a^2-x^2)}dx=\frac{d}{da}{\int }_0^1{a}^n{t}^n{a}^2\sqrt{(1-t^2)}dt={\int }_0^1\frac{d}{da}\left(a^{n+2}{t}^n\sqrt{(1-t^2}dt\right)=$$$$(n+2){\int }_0^1{a}^{n+1}{t}^n\sqrt{(1-t^2}dt$$$${\int }_0^2{x}^3\sqrt{(4-x^2)}dx$$$$n=3a=2$$$$=2^5(I1-I3)=2^5(1-\frac{2}{3})=\frac{32}{3}$$

Commented by maxmathsup by imad last updated on 31/May/19

$$thankssir.$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin