Find 2 equations from the given graph to get the 2 unknowns
0
$begingroup$
The question is:
The curve $(C')$ in the above figure is the represantative curve of the function $f$ defined over $]0,+infty[$ by $f(x)=frac{a ln x}{b+x}$.
Find $a$ and $b$ knowing that $x=0$ is an asymptote to $(C')$ and that the area of the domain limited by $(C')$, the $x$-axis and the straight line of equation $x=2$ is $A=frac{(ln 2)^2}{2}$
I found 1 of the equations from the fact that $f'(1)=1$
But I tried to use the area and the curve and I didn't know how to integrate this function $f(x)$ from $x=1$ to $x=2$
And the asymptotes also didn't help and neither did the fact that $f(1)=0$
Can anyone help with the integral please? Or give another possible solution?
calculus integration indefinite-integrals
edited Dec 18 '18 at 20:34
user376343
3,8133828
asked Dec 16 '18 at 15:07
Fareed AFFareed AF
52612
$endgroup$
add a comment |
0
$begingroup$
The question is:
The curve $(C')$ in the above figure is the represantative curve of the function $f$ defined over $]0,+infty[$ by $f(x)=frac{a ln x}{b+x}$.
Find $a$ and $b$ knowing that $x=0$ is an asymptote to $(C')$ and that the area of the domain limited by $(C')$, the $x$-axis and the straight line of equation $x=2$ is $A=frac{(ln 2)^2}{2}$
I found 1 of the equations from the fact that $f'(1)=1$
But I tried to use the area and the curve and I didn't know how to integrate this function $f(x)$ from $x=1$ to $x=2$
And the asymptotes also didn't help and neither did the fact that $f(1)=0$
Can anyone help with the integral please? Or give another possible solution?
calculus integration indefinite-integrals
edited Dec 18 '18 at 20:34
user376343
3,8133828
asked Dec 16 '18 at 15:07
Fareed AFFareed AF
52612
$endgroup$
$begingroup$
I am sorry, I will do that now
$endgroup$
– Fareed AF
Dec 17 '18 at 4:18
$begingroup$
From the graph it seems that $f$ has a maximum at $x=2.$ From this you can get $b,$ then from $f'(1)$ you get $a$.
$endgroup$
– user376343
Dec 17 '18 at 10:25
$begingroup$
It seems so but its not true that it has a maximum at $2$. And usually if the maximum will be used they put it in the given or they show it clearly in the graph, but here they have given us the area so I think we must use it
$endgroup$
– Fareed AF
Dec 17 '18 at 12:57
$begingroup$
What is xx' ? I don't know this notation.
$endgroup$
– user376343
Dec 17 '18 at 19:42
$begingroup$
The x-axis or the axis of abscissas
$endgroup$
– Fareed AF
Dec 18 '18 at 6:53
add a comment |
0
0
0
$begingroup$
The question is:
The curve $(C')$ in the above figure is the represantative curve of the function $f$ defined over $]0,+infty[$ by $f(x)=frac{a ln x}{b+x}$.
Find $a$ and $b$ knowing that $x=0$ is an asymptote to $(C')$ and that the area of the domain limited by $(C')$, the $x$-axis and the straight line of equation $x=2$ is $A=frac{(ln 2)^2}{2}$
I found 1 of the equations from the fact that $f'(1)=1$
But I tried to use the area and the curve and I didn't know how to integrate this function $f(x)$ from $x=1$ to $x=2$
And the asymptotes also didn't help and neither did the fact that $f(1)=0$
Can anyone help with the integral please? Or give another possible solution?
calculus integration indefinite-integrals
edited Dec 18 '18 at 20:34
user376343
3,8133828
asked Dec 16 '18 at 15:07
Fareed AFFareed AF
52612
$endgroup$
The question is:
The curve $(C')$ in the above figure is the represantative curve of the function $f$ defined over $]0,+infty[$ by $f(x)=frac{a ln x}{b+x}$.
Find $a$ and $b$ knowing that $x=0$ is an asymptote to $(C')$ and that the area of the domain limited by $(C')$, the $x$-axis and the straight line of equation $x=2$ is $A=frac{(ln 2)^2}{2}$
I found 1 of the equations from the fact that $f'(1)=1$
But I tried to use the area and the curve and I didn't know how to integrate this function $f(x)$ from $x=1$ to $x=2$
And the asymptotes also didn't help and neither did the fact that $f(1)=0$
Can anyone help with the integral please? Or give another possible solution?
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Dec 18 '18 at 20:34
user376343
3,8133828
asked Dec 16 '18 at 15:07
Fareed AFFareed AF
52612
edited Dec 18 '18 at 20:34
user376343
3,8133828
asked Dec 16 '18 at 15:07
Fareed AFFareed AF
52612
edited Dec 18 '18 at 20:34
user376343
3,8133828
edited Dec 18 '18 at 20:34
user376343
3,8133828
edited Dec 18 '18 at 20:34
user376343
3,8133828
3,8133828
asked Dec 16 '18 at 15:07
Fareed AFFareed AF
52612
asked Dec 16 '18 at 15:07
Fareed AFFareed AF
52612
asked Dec 16 '18 at 15:07
Fareed AFFareed AF
52612
52612
$begingroup$
I am sorry, I will do that now
$endgroup$
– Fareed AF
Dec 17 '18 at 4:18
$begingroup$
From the graph it seems that $f$ has a maximum at $x=2.$ From this you can get $b,$ then from $f'(1)$ you get $a$.
$endgroup$
– user376343
Dec 17 '18 at 10:25
$begingroup$
It seems so but its not true that it has a maximum at $2$. And usually if the maximum will be used they put it in the given or they show it clearly in the graph, but here they have given us the area so I think we must use it
$endgroup$
– Fareed AF
Dec 17 '18 at 12:57
$begingroup$
What is xx' ? I don't know this notation.
$endgroup$
– user376343
Dec 17 '18 at 19:42
$begingroup$
The x-axis or the axis of abscissas
$endgroup$
– Fareed AF
Dec 18 '18 at 6:53
add a comment |
$begingroup$
I am sorry, I will do that now
$endgroup$
– Fareed AF
Dec 17 '18 at 4:18
$begingroup$
From the graph it seems that $f$ has a maximum at $x=2.$ From this you can get $b,$ then from $f'(1)$ you get $a$.
$endgroup$
– user376343
Dec 17 '18 at 10:25
$begingroup$
It seems so but its not true that it has a maximum at $2$. And usually if the maximum will be used they put it in the given or they show it clearly in the graph, but here they have given us the area so I think we must use it
$endgroup$
– Fareed AF
Dec 17 '18 at 12:57
$begingroup$
What is xx' ? I don't know this notation.
$endgroup$
– user376343
Dec 17 '18 at 19:42
$begingroup$
The x-axis or the axis of abscissas
$endgroup$
– Fareed AF
Dec 18 '18 at 6:53
$begingroup$
I am sorry, I will do that now
$endgroup$
– Fareed AF
Dec 17 '18 at 4:18
$begingroup$
I am sorry, I will do that now
$endgroup$
– Fareed AF
Dec 17 '18 at 4:18
$begingroup$
From the graph it seems that $f$ has a maximum at $x=2.$ From this you can get $b,$ then from $f'(1)$ you get $a$.
$endgroup$
– user376343
Dec 17 '18 at 10:25
$begingroup$
From the graph it seems that $f$ has a maximum at $x=2.$ From this you can get $b,$ then from $f'(1)$ you get $a$.
$endgroup$
– user376343
Dec 17 '18 at 10:25
$begingroup$
It seems so but its not true that it has a maximum at $2$. And usually if the maximum will be used they put it in the given or they show it clearly in the graph, but here they have given us the area so I think we must use it
$endgroup$
– Fareed AF
Dec 17 '18 at 12:57
$begingroup$
It seems so but its not true that it has a maximum at $2$. And usually if the maximum will be used they put it in the given or they show it clearly in the graph, but here they have given us the area so I think we must use it
$endgroup$
– Fareed AF
Dec 17 '18 at 12:57
$begingroup$
What is xx' ? I don't know this notation.
$endgroup$
– user376343
Dec 17 '18 at 19:42
$begingroup$
What is xx' ? I don't know this notation.
$endgroup$
– user376343
Dec 17 '18 at 19:42
$begingroup$
The x-axis or the axis of abscissas
$endgroup$
– Fareed AF
Dec 18 '18 at 6:53
$begingroup$
The x-axis or the axis of abscissas
$endgroup$
– Fareed AF
Dec 18 '18 at 6:53
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Resume:
- As $x=0$ is an asymptote to $(C^{'}),$ necessarilly $bgeq0.$
- From $1=f'(1)=frac{a}{b+1}$ we deduce $a=b+1.$
The function $g(x)=frac{ln x}{x}$ satisfies all constraints, because $int_1^2 frac{ln x}{x} dx =frac 12 (ln 2)^2=mathcal{A}.$ Here $b=0, a=1.$
But $$frac{aln x}{b+x}-frac{ln x}{x}=frac{(x-1)(a-1)ln x}{x(b+x)}>0 quadtext{for}; xin (1,2]quadtext{and}quad ;a=b+1> 1.$$
Therefore we have $$int_1^2 frac{aln x}{b+x} dx >frac 12 (ln 2)^2 quad text{if}quad b>0.$$ The only convenient $b$ is $0$, and $a=1.$
answered Dec 18 '18 at 20:34
user376343user376343
3,8133828
$endgroup$
$begingroup$
I appreciate your help @user376343 I didn't quite understand what you want me to conclude at the end. Also this method of solving is like a trial and error way, can you try to find a more mathematical way to solve this? I mean without expecting what the right answer will be, and just getting it through some derivations and equations ...
$endgroup$
– Fareed AF
Dec 18 '18 at 20:49
$begingroup$
This is surely not a trial - error. A function of given form with $bneq0$ has a polylogarithm in the integral ... far from being probable in this kind of exercise. Thus it was reasonable to check the limit case $b=0.$ And then to find an argument for why it must be this function. This argument looks trivial now, when done :)
$endgroup$
– user376343
Dec 18 '18 at 23:36
$begingroup$
You mean that if they didn't put any given about the area, the exercise can still be solved? They usually only give the needed given to solve the exercise and not more, thats why I think that there must be another solution. I am thankful again and I am sorry for the many questions I am asking but can you please explain a bit more how you found out that $b=0$?
$endgroup$
– Fareed AF
Dec 19 '18 at 4:40
$begingroup$
The area was important. Without, it would not be possible to exclude all possibilities except the one $b=0.$ I justified that in the case $b=0$ takes the area the given value, but in all other cases it is larger.
$endgroup$
– user376343
Dec 19 '18 at 7:31
1
$begingroup$
The integral is larger because the function with $b=1$ takes larger value than $frac{ln x}{x}$ for every $xin (1,2],$ as I proved in my answer. [refer to en.wikipedia.org/wiki/Integral#Inequalities ]
$endgroup$
– user376343
Dec 20 '18 at 21:43
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Resume:
- As $x=0$ is an asymptote to $(C^{'}),$ necessarilly $bgeq0.$
- From $1=f'(1)=frac{a}{b+1}$ we deduce $a=b+1.$
The function $g(x)=frac{ln x}{x}$ satisfies all constraints, because $int_1^2 frac{ln x}{x} dx =frac 12 (ln 2)^2=mathcal{A}.$ Here $b=0, a=1.$
But $$frac{aln x}{b+x}-frac{ln x}{x}=frac{(x-1)(a-1)ln x}{x(b+x)}>0 quadtext{for}; xin (1,2]quadtext{and}quad ;a=b+1> 1.$$
Therefore we have $$int_1^2 frac{aln x}{b+x} dx >frac 12 (ln 2)^2 quad text{if}quad b>0.$$ The only convenient $b$ is $0$, and $a=1.$
answered Dec 18 '18 at 20:34
user376343user376343
3,8133828
$endgroup$
$begingroup$
I appreciate your help @user376343 I didn't quite understand what you want me to conclude at the end. Also this method of solving is like a trial and error way, can you try to find a more mathematical way to solve this? I mean without expecting what the right answer will be, and just getting it through some derivations and equations ...
$endgroup$
– Fareed AF
Dec 18 '18 at 20:49
$begingroup$
This is surely not a trial - error. A function of given form with $bneq0$ has a polylogarithm in the integral ... far from being probable in this kind of exercise. Thus it was reasonable to check the limit case $b=0.$ And then to find an argument for why it must be this function. This argument looks trivial now, when done :)
$endgroup$
– user376343
Dec 18 '18 at 23:36
$begingroup$
You mean that if they didn't put any given about the area, the exercise can still be solved? They usually only give the needed given to solve the exercise and not more, thats why I think that there must be another solution. I am thankful again and I am sorry for the many questions I am asking but can you please explain a bit more how you found out that $b=0$?
$endgroup$
– Fareed AF
Dec 19 '18 at 4:40
$begingroup$
The area was important. Without, it would not be possible to exclude all possibilities except the one $b=0.$ I justified that in the case $b=0$ takes the area the given value, but in all other cases it is larger.
$endgroup$
– user376343
Dec 19 '18 at 7:31
1
$begingroup$
The integral is larger because the function with $b=1$ takes larger value than $frac{ln x}{x}$ for every $xin (1,2],$ as I proved in my answer. [refer to en.wikipedia.org/wiki/Integral#Inequalities ]
$endgroup$
– user376343
Dec 20 '18 at 21:43
|
show 2 more comments
1
$begingroup$
Resume:
- As $x=0$ is an asymptote to $(C^{'}),$ necessarilly $bgeq0.$
- From $1=f'(1)=frac{a}{b+1}$ we deduce $a=b+1.$
The function $g(x)=frac{ln x}{x}$ satisfies all constraints, because $int_1^2 frac{ln x}{x} dx =frac 12 (ln 2)^2=mathcal{A}.$ Here $b=0, a=1.$
But $$frac{aln x}{b+x}-frac{ln x}{x}=frac{(x-1)(a-1)ln x}{x(b+x)}>0 quadtext{for}; xin (1,2]quadtext{and}quad ;a=b+1> 1.$$
Therefore we have $$int_1^2 frac{aln x}{b+x} dx >frac 12 (ln 2)^2 quad text{if}quad b>0.$$ The only convenient $b$ is $0$, and $a=1.$
answered Dec 18 '18 at 20:34
user376343user376343
3,8133828
$endgroup$
$begingroup$
I appreciate your help @user376343 I didn't quite understand what you want me to conclude at the end. Also this method of solving is like a trial and error way, can you try to find a more mathematical way to solve this? I mean without expecting what the right answer will be, and just getting it through some derivations and equations ...
$endgroup$
– Fareed AF
Dec 18 '18 at 20:49
$begingroup$
This is surely not a trial - error. A function of given form with $bneq0$ has a polylogarithm in the integral ... far from being probable in this kind of exercise. Thus it was reasonable to check the limit case $b=0.$ And then to find an argument for why it must be this function. This argument looks trivial now, when done :)
$endgroup$
– user376343
Dec 18 '18 at 23:36
$begingroup$
You mean that if they didn't put any given about the area, the exercise can still be solved? They usually only give the needed given to solve the exercise and not more, thats why I think that there must be another solution. I am thankful again and I am sorry for the many questions I am asking but can you please explain a bit more how you found out that $b=0$?
$endgroup$
– Fareed AF
Dec 19 '18 at 4:40
$begingroup$
The area was important. Without, it would not be possible to exclude all possibilities except the one $b=0.$ I justified that in the case $b=0$ takes the area the given value, but in all other cases it is larger.
$endgroup$
– user376343
Dec 19 '18 at 7:31
1
$begingroup$
The integral is larger because the function with $b=1$ takes larger value than $frac{ln x}{x}$ for every $xin (1,2],$ as I proved in my answer. [refer to en.wikipedia.org/wiki/Integral#Inequalities ]
$endgroup$
– user376343
Dec 20 '18 at 21:43
|
show 2 more comments
1
1
1
$begingroup$
Resume:
- As $x=0$ is an asymptote to $(C^{'}),$ necessarilly $bgeq0.$
- From $1=f'(1)=frac{a}{b+1}$ we deduce $a=b+1.$
The function $g(x)=frac{ln x}{x}$ satisfies all constraints, because $int_1^2 frac{ln x}{x} dx =frac 12 (ln 2)^2=mathcal{A}.$ Here $b=0, a=1.$
But $$frac{aln x}{b+x}-frac{ln x}{x}=frac{(x-1)(a-1)ln x}{x(b+x)}>0 quadtext{for}; xin (1,2]quadtext{and}quad ;a=b+1> 1.$$
Therefore we have $$int_1^2 frac{aln x}{b+x} dx >frac 12 (ln 2)^2 quad text{if}quad b>0.$$ The only convenient $b$ is $0$, and $a=1.$
answered Dec 18 '18 at 20:34
user376343user376343
3,8133828
$endgroup$
Resume:
- As $x=0$ is an asymptote to $(C^{'}),$ necessarilly $bgeq0.$
- From $1=f'(1)=frac{a}{b+1}$ we deduce $a=b+1.$
The function $g(x)=frac{ln x}{x}$ satisfies all constraints, because $int_1^2 frac{ln x}{x} dx =frac 12 (ln 2)^2=mathcal{A}.$ Here $b=0, a=1.$
But $$frac{aln x}{b+x}-frac{ln x}{x}=frac{(x-1)(a-1)ln x}{x(b+x)}>0 quadtext{for}; xin (1,2]quadtext{and}quad ;a=b+1> 1.$$
Therefore we have $$int_1^2 frac{aln x}{b+x} dx >frac 12 (ln 2)^2 quad text{if}quad b>0.$$ The only convenient $b$ is $0$, and $a=1.$
answered Dec 18 '18 at 20:34
user376343user376343
3,8133828
edited Dec 19 '18 at 0:43
answered Dec 18 '18 at 20:34
user376343user376343
3,8133828
answered Dec 18 '18 at 20:34
user376343user376343
3,8133828
answered Dec 18 '18 at 20:34
user376343user376343
3,8133828
3,8133828
$begingroup$
I appreciate your help @user376343 I didn't quite understand what you want me to conclude at the end. Also this method of solving is like a trial and error way, can you try to find a more mathematical way to solve this? I mean without expecting what the right answer will be, and just getting it through some derivations and equations ...
$endgroup$
– Fareed AF
Dec 18 '18 at 20:49
$begingroup$
This is surely not a trial - error. A function of given form with $bneq0$ has a polylogarithm in the integral ... far from being probable in this kind of exercise. Thus it was reasonable to check the limit case $b=0.$ And then to find an argument for why it must be this function. This argument looks trivial now, when done :)
$endgroup$
– user376343
Dec 18 '18 at 23:36
$begingroup$
You mean that if they didn't put any given about the area, the exercise can still be solved? They usually only give the needed given to solve the exercise and not more, thats why I think that there must be another solution. I am thankful again and I am sorry for the many questions I am asking but can you please explain a bit more how you found out that $b=0$?
$endgroup$
– Fareed AF
Dec 19 '18 at 4:40
$begingroup$
The area was important. Without, it would not be possible to exclude all possibilities except the one $b=0.$ I justified that in the case $b=0$ takes the area the given value, but in all other cases it is larger.
$endgroup$
– user376343
Dec 19 '18 at 7:31
1
$begingroup$
The integral is larger because the function with $b=1$ takes larger value than $frac{ln x}{x}$ for every $xin (1,2],$ as I proved in my answer. [refer to en.wikipedia.org/wiki/Integral#Inequalities ]
$endgroup$
– user376343
Dec 20 '18 at 21:43
|
show 2 more comments
$begingroup$
I appreciate your help @user376343 I didn't quite understand what you want me to conclude at the end. Also this method of solving is like a trial and error way, can you try to find a more mathematical way to solve this? I mean without expecting what the right answer will be, and just getting it through some derivations and equations ...
$endgroup$
– Fareed AF
Dec 18 '18 at 20:49
$begingroup$
This is surely not a trial - error. A function of given form with $bneq0$ has a polylogarithm in the integral ... far from being probable in this kind of exercise. Thus it was reasonable to check the limit case $b=0.$ And then to find an argument for why it must be this function. This argument looks trivial now, when done :)
$endgroup$
– user376343
Dec 18 '18 at 23:36
$begingroup$
You mean that if they didn't put any given about the area, the exercise can still be solved? They usually only give the needed given to solve the exercise and not more, thats why I think that there must be another solution. I am thankful again and I am sorry for the many questions I am asking but can you please explain a bit more how you found out that $b=0$?
$endgroup$
– Fareed AF
Dec 19 '18 at 4:40
$begingroup$
The area was important. Without, it would not be possible to exclude all possibilities except the one $b=0.$ I justified that in the case $b=0$ takes the area the given value, but in all other cases it is larger.
$endgroup$
– user376343
Dec 19 '18 at 7:31
1
$begingroup$
The integral is larger because the function with $b=1$ takes larger value than $frac{ln x}{x}$ for every $xin (1,2],$ as I proved in my answer. [refer to en.wikipedia.org/wiki/Integral#Inequalities ]
$endgroup$
– user376343
Dec 20 '18 at 21:43
$begingroup$
I appreciate your help @user376343 I didn't quite understand what you want me to conclude at the end. Also this method of solving is like a trial and error way, can you try to find a more mathematical way to solve this? I mean without expecting what the right answer will be, and just getting it through some derivations and equations ...
$endgroup$
– Fareed AF
Dec 18 '18 at 20:49
$begingroup$
I appreciate your help @user376343 I didn't quite understand what you want me to conclude at the end. Also this method of solving is like a trial and error way, can you try to find a more mathematical way to solve this? I mean without expecting what the right answer will be, and just getting it through some derivations and equations ...
$endgroup$
– Fareed AF
Dec 18 '18 at 20:49
$begingroup$
This is surely not a trial - error. A function of given form with $bneq0$ has a polylogarithm in the integral ... far from being probable in this kind of exercise. Thus it was reasonable to check the limit case $b=0.$ And then to find an argument for why it must be this function. This argument looks trivial now, when done :)
$endgroup$
– user376343
Dec 18 '18 at 23:36
$begingroup$
This is surely not a trial - error. A function of given form with $bneq0$ has a polylogarithm in the integral ... far from being probable in this kind of exercise. Thus it was reasonable to check the limit case $b=0.$ And then to find an argument for why it must be this function. This argument looks trivial now, when done :)
$endgroup$
– user376343
Dec 18 '18 at 23:36
$begingroup$
You mean that if they didn't put any given about the area, the exercise can still be solved? They usually only give the needed given to solve the exercise and not more, thats why I think that there must be another solution. I am thankful again and I am sorry for the many questions I am asking but can you please explain a bit more how you found out that $b=0$?
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– Fareed AF
Dec 19 '18 at 4:40
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You mean that if they didn't put any given about the area, the exercise can still be solved? They usually only give the needed given to solve the exercise and not more, thats why I think that there must be another solution. I am thankful again and I am sorry for the many questions I am asking but can you please explain a bit more how you found out that $b=0$?
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– Fareed AF
Dec 19 '18 at 4:40
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The area was important. Without, it would not be possible to exclude all possibilities except the one $b=0.$ I justified that in the case $b=0$ takes the area the given value, but in all other cases it is larger.
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– user376343
Dec 19 '18 at 7:31
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The area was important. Without, it would not be possible to exclude all possibilities except the one $b=0.$ I justified that in the case $b=0$ takes the area the given value, but in all other cases it is larger.
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– user376343
Dec 19 '18 at 7:31
1
1
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The integral is larger because the function with $b=1$ takes larger value than $frac{ln x}{x}$ for every $xin (1,2],$ as I proved in my answer. [refer to en.wikipedia.org/wiki/Integral#Inequalities ]
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– user376343
Dec 20 '18 at 21:43
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The integral is larger because the function with $b=1$ takes larger value than $frac{ln x}{x}$ for every $xin (1,2],$ as I proved in my answer. [refer to en.wikipedia.org/wiki/Integral#Inequalities ]
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– user376343
Dec 20 '18 at 21:43
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I am sorry, I will do that now
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– Fareed AF
Dec 17 '18 at 4:18
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From the graph it seems that $f$ has a maximum at $x=2.$ From this you can get $b,$ then from $f'(1)$ you get $a$.
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– user376343
Dec 17 '18 at 10:25
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It seems so but its not true that it has a maximum at $2$. And usually if the maximum will be used they put it in the given or they show it clearly in the graph, but here they have given us the area so I think we must use it
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– Fareed AF
Dec 17 '18 at 12:57
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What is xx' ? I don't know this notation.
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– user376343
Dec 17 '18 at 19:42
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The x-axis or the axis of abscissas
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– Fareed AF
Dec 18 '18 at 6:53