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Compute-lim-A-1-A-1-A-A-1-x-dx-IMC-2015-

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Question Number 7501 by Yozzia last updated on 01/Sep/16
Compute lim_(A→+∞) {(1/A)∫_1 ^A A^(1/x) dx}. (IMC 2015)
$${Compute}\:\underset{{A}\rightarrow+\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{A}}\int_{\mathrm{1}} ^{{A}} {A}^{\frac{\mathrm{1}}{{x}}} {dx}\right\}. \\ $$$$\left({IMC}\:\mathrm{2015}\right) \\ $$
Answered by FilupSmith last updated on 01/Sep/16
L =lim_(A→∞) {(1/A)∫_1 ^A A^(1/x) dx} unsure if math is correct. L =lim_(A→∞) {(1/A)∫_1 ^A e^((1/x)ln(A)) dx} u=e^((1/x)ln(A)) ⇒ du=−x^(−2) ln(A)e^(x^(−1) ln(A)) dx x=((ln(A))/(ln(u))) du=−((ln(u))/x)udx du=−((ln^2 (u))/(ln(A)))udx dx=−((ln(A))/(uln^2 (u)))du L =lim_(A→∞) {(1/A)∫_(x=1) ^( x=A) u(−((ln(A))/(uln^2 (u)))du)} L =lim_(A→∞) {−(1/A)∫_(x=1) ^( x=A) u((ln(A))/(uln^2 (u)))du} L =lim_(A→∞) {−((ln(A))/A)∫_(x=1) ^( x=A) (1/(ln^2 (u)))du} limit of ((ln(A))/A)=(1/A) as A→∞ = 0 0×constant=0 continue
$${L}\:=\underset{{A}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{A}}\int_{\mathrm{1}} ^{{A}} {A}^{\frac{\mathrm{1}}{{x}}} {dx}\right\} \\ $$$${unsure}\:{if}\:{math}\:{is}\:{correct}. \\ $$$${L}\:=\underset{{A}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{A}}\int_{\mathrm{1}} ^{{A}} {e}^{\frac{\mathrm{1}}{{x}}\mathrm{ln}\left({A}\right)} {dx}\right\} \\ $$$${u}={e}^{\frac{\mathrm{1}}{{x}}\mathrm{ln}\left({A}\right)} \:\Rightarrow\:{du}=−{x}^{−\mathrm{2}} \mathrm{ln}\left({A}\right){e}^{{x}^{−\mathrm{1}} \mathrm{ln}\left({A}\right)} {dx} \\ $$$${x}=\frac{\mathrm{ln}\left({A}\right)}{\mathrm{ln}\left({u}\right)} \\ $$$${du}=−\frac{\mathrm{ln}\left({u}\right)}{{x}}{udx} \\ $$$${du}=−\frac{\mathrm{ln}^{\mathrm{2}} \left({u}\right)}{\mathrm{ln}\left({A}\right)}{udx} \\ $$$${dx}=−\frac{\mathrm{ln}\left({A}\right)}{{u}\mathrm{ln}^{\mathrm{2}} \left({u}\right)}{du} \\ $$$${L}\:=\underset{{A}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{A}}\int_{{x}=\mathrm{1}} ^{\:{x}={A}} {u}\left(−\frac{\mathrm{ln}\left({A}\right)}{{u}\mathrm{ln}^{\mathrm{2}} \left({u}\right)}{du}\right)\right\} \\ $$$${L}\:=\underset{{A}\rightarrow\infty} {\mathrm{lim}}\left\{−\frac{\mathrm{1}}{{A}}\int_{{x}=\mathrm{1}} ^{\:{x}={A}} {u}\frac{\mathrm{ln}\left({A}\right)}{{u}\mathrm{ln}^{\mathrm{2}} \left({u}\right)}{du}\right\} \\ $$$${L}\:=\underset{{A}\rightarrow\infty} {\mathrm{lim}}\left\{−\frac{\mathrm{ln}\left({A}\right)}{{A}}\int_{{x}=\mathrm{1}} ^{\:{x}={A}} \frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left({u}\right)}{du}\right\} \\ $$$${limit}\:{of}\:\:\frac{\mathrm{ln}\left({A}\right)}{{A}}=\frac{\mathrm{1}}{{A}}\:{as}\:{A}\rightarrow\infty\:=\:\mathrm{0} \\ $$$$\mathrm{0}×{constant}=\mathrm{0} \\ $$$${continue} \\ $$
Commented by FilupSmith last updated on 01/Sep/16
∫(1/(ln^2 (u)))du t=ln^2 (u) u=e^t^(1/2) dt=2ln(u)(1/u)du (u/(2ln(u)))dt=du (e^t^(1/2) /(2t^(1/2) ))dt=du L =lim_(A→∞) {−((ln(A))/A)∫_(x=1) ^( x=A) (1/(ln^2 (u)))du} L =lim_(A→∞) {−((ln(A))/A)∫_(x=1) ^( x=A) (e^t^(1/2) /(2t^(3/2) ))dt} correct???
$$\int\frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left({u}\right)}{du} \\ $$$${t}=\mathrm{ln}^{\mathrm{2}} \left({u}\right) \\ $$$${u}={e}^{{t}^{\mathrm{1}/\mathrm{2}} } \\ $$$${dt}=\mathrm{2ln}\left({u}\right)\frac{\mathrm{1}}{{u}}{du} \\ $$$$\frac{{u}}{\mathrm{2ln}\left({u}\right)}{dt}={du} \\ $$$$\frac{{e}^{{t}^{\mathrm{1}/\mathrm{2}} } }{\mathrm{2}{t}^{\mathrm{1}/\mathrm{2}} }{dt}={du} \\ $$$${L}\:=\underset{{A}\rightarrow\infty} {\mathrm{lim}}\left\{−\frac{\mathrm{ln}\left({A}\right)}{{A}}\int_{{x}=\mathrm{1}} ^{\:{x}={A}} \frac{\mathrm{1}}{\mathrm{ln}^{\mathrm{2}} \left({u}\right)}{du}\right\} \\ $$$${L}\:=\underset{{A}\rightarrow\infty} {\mathrm{lim}}\left\{−\frac{\mathrm{ln}\left({A}\right)}{{A}}\int_{{x}=\mathrm{1}} ^{\:{x}={A}} \frac{{e}^{{t}^{\mathrm{1}/\mathrm{2}} } }{\mathrm{2}{t}^{\mathrm{3}/\mathrm{2}} }{dt}\right\} \\ $$$${correct}??? \\ $$
Commented by Yozzia last updated on 01/Sep/16
So far it seems ok. I can access the solution online, but I don′t know it. It′s a question from the 2015 International Mathematics Competition for Universities. Just recently I got wind of this competition and now I have access to a variety of intriguing problems which I′ll hopefully be able to solve as I progress in my undergraduate studies. I′m hoping to share some of those problems here with the forum!
$${So}\:{far}\:{it}\:{seems}\:{ok}.\:{I}\:{can}\:{access}\:{the} \\ $$$${solution}\:{online},\:{but}\:{I}\:{don}'{t}\:{know}\:{it}. \\ $$$${It}'{s}\:{a}\:{question}\:{from}\:{the}\:\mathrm{2015}\:{International} \\ $$$${Mathematics}\:{Competition}\:{for}\:{Universities}. \\ $$$${Just}\:{recently}\:{I}\:{got}\:{wind}\:{of}\:{this}\:{competition} \\ $$$${and}\:{now}\:{I}\:{have}\:{access}\:{to}\:{a}\:{variety}\:{of} \\ $$$${intriguing}\:{problems}\:{which}\:{I}'{ll}\:{hopefully} \\ $$$${be}\:{able}\:{to}\:{solve}\:{as}\:{I}\:{progress}\:{in}\:{my}\: \\ $$$${undergraduate}\:{studies}.\:{I}'{m}\:{hoping} \\ $$$${to}\:{share}\:{some}\:{of}\:{those}\:{problems}\:{here} \\ $$$${with}\:{the}\:{forum}! \\ $$
Commented by FilupSmith last updated on 02/Sep/16
That is awsome. I have no formal education. I am completely self taught and you have taught me lot!
$$\mathrm{That}\:\mathrm{is}\:\mathrm{awsome}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{formal}\:\mathrm{education}. \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{completely}\:\mathrm{self}\:\mathrm{taught}\:\mathrm{and}\:\mathrm{you}\:\mathrm{have} \\ $$$$\mathrm{taught}\:\mathrm{me}\:\:\mathrm{lot}! \\ $$
Commented by Alimurtaza last updated on 05/Jan/17
$$ \\ $$

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