Question Number 1625 by 112358 last updated on 27/Aug/15
$${y}\left({x}\right)=\frac{\mathrm{1}}{{x}−{a}}\int_{{a}} ^{{x}} \sqrt{{t}+\sqrt{{t}+\sqrt{{t}+\sqrt{{t}+\sqrt{{t}+…}}}}}{dt} \\ $$$${x}\neq{a},\:{a}>\mathrm{0},{y}\left({x}\right)>\mathrm{0}. \\ $$$${Find}\:\:{y}\left(\mathrm{2}{a}\right). \\ $$
Answered by Rasheed Soomro last updated on 28/Aug/15
![y(x)=(1/(x−a))∫_a ^( x) ( (√(t+(√(t+(√(t+(√(t+(√(t+...)))))))))) )dt Let (√(t+(√(t+(√(t+(√(t+(√(t+...))))))))))=u Squaring both sides: t+(√(t+(√(t+(√(t+(√(t+...))))))))=u^2 t+u=u^(2 ) ⇒ u^2 −u−t=0 u=((1±(√(1+4t)))/2) y(x)=(1/(x−a)) ∫_a ^( x) (((1±(√(1+4t)))/2) )dt =(1/(2(x−a)))∫_a ^( x) (1±(√(1+4t)))dt =(1/(2(x−a)))[t±(1/6)(1+4t)^(3/2) ]_a ^x =(1/(2(x−a)))([x±((√((1+4x)^3 ))/6)]−[a±((√((1+4a)^3 ))/6)]) =(1/(2(x−a)))×((6(x−a)±(√((1+4x)^3 )) ∓(√((1+4a)^3 )) )/6) =((6(x−a)±((√((1+4x)^3 )) −(√((1+4a)^3 )) ))/(12(x−a))) y(x)=(1/2)±(((√((1+4x)^3 )) −(√((1+4a)^3 )))/(12(x−a))) y(2a)=(1/2)±(((√((1+4(2a))^3 )) −(√((1+4a)^3 )))/(12((2a)−a))) =(1/2)±(((√((1+8a)^3 )) −(√((1+4a)^3 )))/(12a))](https://www.tinkutara.com/question/Q1630.png)
$${y}\left({x}\right)=\frac{\mathrm{1}}{{x}−{a}}\int_{{a}} ^{\:{x}} \left(\:\sqrt{{t}+\sqrt{{t}+\sqrt{{t}+\sqrt{{t}+\sqrt{{t}+…}}}}}\:\right){dt} \\ $$$${Let}\:\:\:\sqrt{{t}+\sqrt{{t}+\sqrt{{t}+\sqrt{{t}+\sqrt{{t}+…}}}}}={u} \\ $$$${Squaring}\:{both}\:{sides}: \\ $$$$\:\:\:\:\:\:\:\:\:\:{t}+\sqrt{{t}+\sqrt{{t}+\sqrt{{t}+\sqrt{{t}+…}}}}={u}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:{t}+{u}={u}^{\mathrm{2}\:\:\:} \Rightarrow\:\:{u}^{\mathrm{2}} −{u}−{t}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{u}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{t}}}{\mathrm{2}} \\ $$$${y}\left({x}\right)=\frac{\mathrm{1}}{{x}−{a}}\:\int_{{a}} ^{\:{x}} \left(\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{t}}}{\mathrm{2}}\:\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\left({x}−{a}\right)}\int_{{a}} ^{\:{x}} \left(\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{t}}\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\left({x}−{a}\right)}\left[{t}\pm\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{1}+\mathrm{4}{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\right]_{{a}} ^{{x}} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}\left({x}−{a}\right)}\left(\left[{x}\pm\frac{\sqrt{\left(\mathrm{1}+\mathrm{4}{x}\right)^{\mathrm{3}} }}{\mathrm{6}}\right]−\left[{a}\pm\frac{\sqrt{\left(\mathrm{1}+\mathrm{4}{a}\right)^{\mathrm{3}} }}{\mathrm{6}}\right]\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({x}−{a}\right)}×\frac{\mathrm{6}\left({x}−{a}\right)\pm\sqrt{\left(\mathrm{1}+\mathrm{4}{x}\right)^{\mathrm{3}} }\:\mp\sqrt{\left(\mathrm{1}+\mathrm{4}{a}\right)^{\mathrm{3}} }\:}{\mathrm{6}} \\ $$$$=\frac{\mathrm{6}\left({x}−{a}\right)\pm\left(\sqrt{\left(\mathrm{1}+\mathrm{4}{x}\right)^{\mathrm{3}} }\:−\sqrt{\left(\mathrm{1}+\mathrm{4}{a}\right)^{\mathrm{3}} }\:\right)}{\mathrm{12}\left({x}−{a}\right)} \\ $$$${y}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\left(\mathrm{1}+\mathrm{4}{x}\right)^{\mathrm{3}} }\:−\sqrt{\left(\mathrm{1}+\mathrm{4}{a}\right)^{\mathrm{3}} }}{\mathrm{12}\left({x}−{a}\right)} \\ $$$${y}\left(\mathrm{2}{a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\left(\mathrm{1}+\mathrm{4}\left(\mathrm{2}{a}\right)\right)^{\mathrm{3}} }\:−\sqrt{\left(\mathrm{1}+\mathrm{4}{a}\right)^{\mathrm{3}} }}{\mathrm{12}\left(\left(\mathrm{2}{a}\right)−{a}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\left(\mathrm{1}+\mathrm{8}{a}\right)^{\mathrm{3}} }\:−\sqrt{\left(\mathrm{1}+\mathrm{4}{a}\right)^{\mathrm{3}} }}{\mathrm{12}{a}} \\ $$
Commented by 123456 last updated on 28/Aug/15
![∫_a ^x 1±(√(1+4t))dt=∫_a ^x dt±∫_a ^x (√(1+4t))dt ∫_a ^x dt=x−a ∫_a ^x (√(1+4t))dt u=1+4t⇒du=4dt t=a⇒u=1+4a t=x⇒u=1+4x ∫_a ^x (√(1+4t))dt=(1/4)∫_(1+4a) ^(1+4x) (√u)du=[((1/4)/(3/2))(√u^3 )]_(1+4a) ^(1+4x) =(((√((1+4x)^3 ))−(√((1+4a)^3 )))/6) ∫_a ^x 1±(√(1+4t))dt=x−a±(((√((1+4x)^3 ))−(√((1+4a)^3 )))/6)](https://www.tinkutara.com/question/Q1637.png)
$$\underset{{a}} {\overset{{x}} {\int}}\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{t}}{dt}=\underset{{a}} {\overset{{x}} {\int}}{dt}\pm\underset{{a}} {\overset{{x}} {\int}}\sqrt{\mathrm{1}+\mathrm{4}{t}}{dt} \\ $$$$\underset{{a}} {\overset{{x}} {\int}}{dt}={x}−{a} \\ $$$$\underset{{a}} {\overset{{x}} {\int}}\sqrt{\mathrm{1}+\mathrm{4}{t}}{dt} \\ $$$${u}=\mathrm{1}+\mathrm{4}{t}\Rightarrow{du}=\mathrm{4}{dt} \\ $$$${t}={a}\Rightarrow{u}=\mathrm{1}+\mathrm{4}{a} \\ $$$${t}={x}\Rightarrow{u}=\mathrm{1}+\mathrm{4}{x} \\ $$$$\underset{{a}} {\overset{{x}} {\int}}\sqrt{\mathrm{1}+\mathrm{4}{t}}{dt}=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{1}+\mathrm{4}{a}} {\overset{\mathrm{1}+\mathrm{4}{x}} {\int}}\sqrt{{u}}{du}=\left[\frac{\mathrm{1}/\mathrm{4}}{\mathrm{3}/\mathrm{2}}\sqrt{{u}^{\mathrm{3}} }\right]_{\mathrm{1}+\mathrm{4}{a}} ^{\mathrm{1}+\mathrm{4}{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\left(\mathrm{1}+\mathrm{4}{x}\right)^{\mathrm{3}} }−\sqrt{\left(\mathrm{1}+\mathrm{4}{a}\right)^{\mathrm{3}} }}{\mathrm{6}} \\ $$$$\underset{{a}} {\overset{{x}} {\int}}\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{t}}{dt}={x}−{a}\pm\frac{\sqrt{\left(\mathrm{1}+\mathrm{4}{x}\right)^{\mathrm{3}} }−\sqrt{\left(\mathrm{1}+\mathrm{4}{a}\right)^{\mathrm{3}} }}{\mathrm{6}} \\ $$
Commented by 123456 last updated on 28/Aug/15
$${y}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}\left({x}−{a}\right)}×\frac{\mathrm{6}\left({x}−{a}\right)\pm\sqrt{\left(\mathrm{1}+\mathrm{4}{x}\right)^{\mathrm{3}} }\mp\sqrt{\left(\mathrm{1}+\mathrm{4}{a}\right)^{\mathrm{3}} }}{\mathrm{6}} \\ $$
Commented by Rasheed Soomro last updated on 28/Aug/15
$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{mention}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{calculation}}\:\boldsymbol{\mathrm{mistake}}.\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{am}}\:\boldsymbol{\mathrm{going}}\:\boldsymbol{\mathrm{to}} \\ $$$$\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{it}}. \\ $$
Commented by 123456 last updated on 28/Aug/15
![lim_(x→a) y(x) is like [(d/dx)∫f(x)dx]_(x=a) :) lim_(x→a) ((∫_a ^x f(x)dx)/(x−a))=lim_(x→a) ((F(x)−F(a))/(x−a))=f(a)](https://www.tinkutara.com/question/Q1647.png)
$$\left.\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{y}\left({x}\right)\:\mathrm{is}\:\mathrm{like}\:\left[\frac{\mathrm{d}}{\mathrm{d}{x}}\int{f}\left({x}\right){dx}\right]_{{x}={a}} \::\right) \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{\underset{{a}} {\overset{{x}} {\int}}{f}\left({x}\right){dx}}{{x}−{a}}=\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{\mathrm{F}\left({x}\right)−\mathrm{F}\left({a}\right)}{{x}−{a}}={f}\left({a}\right) \\ $$
Commented by 112358 last updated on 29/Aug/15
![I created this question having the formula for the average value of a function in mind: y_(avg) =(1/(b−a))∫_a ^b f(x)dx . y_(avg) =((F(b)−F(a))/(b−a)) (∗) (∗) resembles the relation given by the mean value theorem which states that ∃c∈[a,b] such that (d/dx)f(x)∣_(x=c) =((f(a)−f(b))/(b−a)) given that f(x) is continuous ∀x∈[a,b] and differentiable ∀x∈(a,b).](https://www.tinkutara.com/question/Q1648.png)
$${I}\:{created}\:{this}\:{question}\:{having}\:{the} \\ $$$${formula}\:{for}\:{the}\:{average}\:{value} \\ $$$${of}\:{a}\:{function}\:{in}\:{mind}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}_{{avg}} =\frac{\mathrm{1}}{{b}−{a}}\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}_{{avg}} =\frac{{F}\left({b}\right)−{F}\left({a}\right)}{{b}−{a}}\:\:\:\:\left(\ast\right) \\ $$$$\left(\ast\right)\:{resembles}\:{the}\:{relation}\:{given} \\ $$$${by}\:{the}\:{mean}\:{value}\:{theorem}\:{which} \\ $$$${states}\:{that}\:\exists{c}\in\left[{a},{b}\right]\:{such}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{d}}{{dx}}{f}\left({x}\right)\mid_{{x}={c}} =\frac{{f}\left({a}\right)−{f}\left({b}\right)}{{b}−{a}} \\ $$$${given}\:{that}\:{f}\left({x}\right)\:{is}\:{continuous}\: \\ $$$$\forall{x}\in\left[{a},{b}\right]\:{and}\:{differentiable} \\ $$$$\forall{x}\in\left({a},{b}\right).\: \\ $$$$ \\ $$$$ \\ $$
Commented by 123456 last updated on 29/Aug/15
![or mean value theorem for integrails if f is continuous and integrable into [a,b] them ∃ξ∈[a,b] ∫_a ^b f(x)dx=f(ξ)(b−a) so for some ξ∈[a,b] y_(avg) =(1/(b−a))∫_a ^b f(x)dx=f(ξ)](https://www.tinkutara.com/question/Q1653.png)
$$\mathrm{or}\:\mathrm{mean}\:\mathrm{value}\:\mathrm{theorem}\:\mathrm{for}\:\mathrm{integrails} \\ $$$$\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{and}\:\mathrm{integrable}\:\mathrm{into}\:\left[{a},{b}\right] \\ $$$$\mathrm{them}\:\exists\xi\in\left[{a},{b}\right] \\ $$$$\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}={f}\left(\xi\right)\left({b}−{a}\right) \\ $$$$\mathrm{so}\:\mathrm{for}\:\mathrm{some}\:\xi\in\left[{a},{b}\right] \\ $$$${y}_{{avg}} =\frac{\mathrm{1}}{{b}−{a}}\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}={f}\left(\xi\right) \\ $$
Commented by 112358 last updated on 29/Aug/15
$${Nice}.\:{Thanks}! \\ $$