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A SL math problem
This was a problem my SL calculus class did for a warm-up. I used one method, my instructor used
another, both were correct, and the answers were different. Now it's extra credit to figure out why
the answers were different, and we're allowed to seek outside help.
Since the symbol for integration can't be used on this message board, I'm going to use S instead. It
will probably help if you write this out on a piece of paper using the real symbols for integrate,
absolute value, and squared.
The problem is as follows:
S (1) / (x ln (x squared)) dx
MY METHOD
S (2) / (x ln (x squared)) dx
S (2/a) / (ln x squared) dx
dx= (1/2) S 2 (absolute value of x) / ln x squared dx
u= ln x squared
u prime= 2/x dx
= (1/2) ln (absolute value of u) + C
=(1/2) ln (absolute value of ln x squared) + C
= (1/2) ln (absolute value of 2 ln x) + C
= ln (absolute value of ln x) + C
MY INSTRUCTORS METHOD
S (3) / (x ln (x squared)) dx
= S (1) / (2x ln x) dx
= (1/2) S (1) / (x ln x) dx
=(1/2) (1/x) (1/ ln x) dx=du
u= ln x
du= (1/x) dx
= (1/2) S (1/u) du
= (1/2) ln (absolute value of u) + C
= (1/2) ln (absolute value of ln x) + C
Now we're posed with the problem of finding out what went wrong. As far as I can see, both methods
were correct, but plugging values in to check the answers and graphing them revealed a graph that
was similar, but slightly off. If anyone has any ideas, PLEASE let me know, I could really use some
extra credit!
Thanks, -Kelley
PS, if anyone wants me to, I can send them an attatchment of the problem that has the correct
symbols that I can't post here.
another, both were correct, and the answers were different. Now it's extra credit to figure out why
the answers were different, and we're allowed to seek outside help.
Since the symbol for integration can't be used on this message board, I'm going to use S instead. It
will probably help if you write this out on a piece of paper using the real symbols for integrate,
absolute value, and squared.
The problem is as follows:
S (1) / (x ln (x squared)) dx
MY METHOD
S (2) / (x ln (x squared)) dx
S (2/a) / (ln x squared) dx
dx= (1/2) S 2 (absolute value of x) / ln x squared dx
u= ln x squared
u prime= 2/x dx
= (1/2) ln (absolute value of u) + C
=(1/2) ln (absolute value of ln x squared) + C
= (1/2) ln (absolute value of 2 ln x) + C
= ln (absolute value of ln x) + C
MY INSTRUCTORS METHOD
S (3) / (x ln (x squared)) dx
= S (1) / (2x ln x) dx
= (1/2) S (1) / (x ln x) dx
=(1/2) (1/x) (1/ ln x) dx=du
u= ln x
du= (1/x) dx
= (1/2) S (1/u) du
= (1/2) ln (absolute value of u) + C
= (1/2) ln (absolute value of ln x) + C
Now we're posed with the problem of finding out what went wrong. As far as I can see, both methods
were correct, but plugging values in to check the answers and graphing them revealed a graph that
was similar, but slightly off. If anyone has any ideas, PLEASE let me know, I could really use some
extra credit!
Thanks, -Kelley
PS, if anyone wants me to, I can send them an attatchment of the problem that has the correct
symbols that I can't post here.
What is "u prime?" I have no idea, I have never heard of the way you did it. Only your instructor's
way. I only did Methods, perhaps an HL student would like to explain. Hmmm... my maths is slowly
going out the window I think...
Cidra17918 wrote:
[q1]> This was a problem my SL calculus class did for a warm-up. I used one method, my instructor used[/q1]
[q1]> another, both were correct, and the answers were different. Now it's extra credit to figure out[/q1]
[q1]> why the answers were different, and we're allowed to seek outside help.[/q1]
[q1]>[/q1]
[q1]> Since the symbol for integration can't be used on this message board, I'm going to use S instead.[/q1]
[q1]> It will probably help if you write this out on a piece of paper using the real symbols for[/q1]
[q1]> integrate, absolute value, and squared.[/q1]
[q1]>[/q1]
[q1]> The problem is as follows:[/q1]
[q1]>[/q1]
[q1]> S (1) / (x ln (x squared)) dx[/q1]
[q1]>[/q1]
[q1]> MY METHOD[/q1]
[q1]>[/q1]
[q1]> S (1) / (x ln (x squared)) dx[/q1]
[q1]>[/q1]
[q1]> S (1/x) / (ln x squared) dx[/q1]
[q1]>[/q1]
[q1]> dx= (1/2) S 2 (absolute value of x) / ln x squared dx[/q1]
[q1]>[/q1]
[q1]> u= ln x squared[/q1]
[q1]>[/q1]
[q1]> u prime= 2/x dx[/q1]
[q1]>[/q1]
Is u prime the derivative of u?
[q1]>[/q1]
[q1]> = (1/2) ln (absolute value of u) + C[/q1]
[q1]>[/q1]
[q1]> =(1/2) ln (absolute value of ln x squared) + C[/q1]
[q1]>[/q1]
[q1]> = (1/2) ln (absolute value of 2 ln x) + C[/q1]
[q1]>[/q1]
[q1]> = ln (absolute value of ln x) + C[/q1]
[q1]>[/q1]
[q1]> MY INSTRUCTORS METHOD[/q1]
[q1]>[/q1]
[q1]> S (1) / (x ln (x squared)) dx[/q1]
[q1]>[/q1]
[q1]> = S (1) / (2x ln x) dx[/q1]
[q1]>[/q1]
This line above uses the fact that you take the x squared on the ln x and put it in front as a
factor of the x.
[q1]>[/q1]
[q1]> = (1/2) S (1) / (x ln x) dx[/q1]
[q1]>[/q1]
Take the two out the front.
[q1]>[/q1]
[q1]> =(1/2) (1/x) (1/ ln x) dx=du[/q1]
[q1]>[/q1]
Put each term separately.
[q1]>[/q1]
[q1]> u= ln x[/q1]
[q1]>[/q1]
Substitution method.
[q1]>[/q1]
[q1]> du= (1/x) dx[/q1]
[q1]>[/q1]
More substitution method.
[q1]>[/q1]
[q1]> = (1/2) S (1/u) du[/q1]
[q1]>[/q1]
Back to original equation using susbtituted values.
[q1]>[/q1]
[q1]> = (1/2) ln (absolute value of u) + C[/q1]
[q1]>[/q1]
Now back to x, substituting back from u.
[q1]>[/q1]
[q1]> = (1/2) ln (absolute value of ln x) + C[/q1]
[q1]>[/q1]
[q1]> Now we're posed with the problem of finding out what went wrong. As far as I can see, both methods[/q1]
[q1]> were correct, but plugging values in to check the answers and graphing them revealed a graph that[/q1]
[q1]> was similar, but slightly off. If anyone has any ideas, PLEASE let me know, I could really use[/q1]
[q1]> some extra credit![/q1]
[q1]>[/q1]
[q1]> Thanks, -Kelley[/q1]
[q1]>[/q1]
[q1]> PS, if anyone wants me to, I can send them an attatchment of the problem that has the correct[/q1]
[q1]> symbols that I can't post here.[/q1]
way. I only did Methods, perhaps an HL student would like to explain. Hmmm... my maths is slowly
going out the window I think...
Cidra17918 wrote:
[q1]> This was a problem my SL calculus class did for a warm-up. I used one method, my instructor used[/q1]
[q1]> another, both were correct, and the answers were different. Now it's extra credit to figure out[/q1]
[q1]> why the answers were different, and we're allowed to seek outside help.[/q1]
[q1]>[/q1]
[q1]> Since the symbol for integration can't be used on this message board, I'm going to use S instead.[/q1]
[q1]> It will probably help if you write this out on a piece of paper using the real symbols for[/q1]
[q1]> integrate, absolute value, and squared.[/q1]
[q1]>[/q1]
[q1]> The problem is as follows:[/q1]
[q1]>[/q1]
[q1]> S (1) / (x ln (x squared)) dx[/q1]
[q1]>[/q1]
[q1]> MY METHOD[/q1]
[q1]>[/q1]
[q1]> S (1) / (x ln (x squared)) dx[/q1]
[q1]>[/q1]
[q1]> S (1/x) / (ln x squared) dx[/q1]
[q1]>[/q1]
[q1]> dx= (1/2) S 2 (absolute value of x) / ln x squared dx[/q1]
[q1]>[/q1]
[q1]> u= ln x squared[/q1]
[q1]>[/q1]
[q1]> u prime= 2/x dx[/q1]
[q1]>[/q1]
Is u prime the derivative of u?
[q1]>[/q1]
[q1]> = (1/2) ln (absolute value of u) + C[/q1]
[q1]>[/q1]
[q1]> =(1/2) ln (absolute value of ln x squared) + C[/q1]
[q1]>[/q1]
[q1]> = (1/2) ln (absolute value of 2 ln x) + C[/q1]
[q1]>[/q1]
[q1]> = ln (absolute value of ln x) + C[/q1]
[q1]>[/q1]
[q1]> MY INSTRUCTORS METHOD[/q1]
[q1]>[/q1]
[q1]> S (1) / (x ln (x squared)) dx[/q1]
[q1]>[/q1]
[q1]> = S (1) / (2x ln x) dx[/q1]
[q1]>[/q1]
This line above uses the fact that you take the x squared on the ln x and put it in front as a
factor of the x.
[q1]>[/q1]
[q1]> = (1/2) S (1) / (x ln x) dx[/q1]
[q1]>[/q1]
Take the two out the front.
[q1]>[/q1]
[q1]> =(1/2) (1/x) (1/ ln x) dx=du[/q1]
[q1]>[/q1]
Put each term separately.
[q1]>[/q1]
[q1]> u= ln x[/q1]
[q1]>[/q1]
Substitution method.
[q1]>[/q1]
[q1]> du= (1/x) dx[/q1]
[q1]>[/q1]
More substitution method.
[q1]>[/q1]
[q1]> = (1/2) S (1/u) du[/q1]
[q1]>[/q1]
Back to original equation using susbtituted values.
[q1]>[/q1]
[q1]> = (1/2) ln (absolute value of u) + C[/q1]
[q1]>[/q1]
Now back to x, substituting back from u.
[q1]>[/q1]
[q1]> = (1/2) ln (absolute value of ln x) + C[/q1]
[q1]>[/q1]
[q1]> Now we're posed with the problem of finding out what went wrong. As far as I can see, both methods[/q1]
[q1]> were correct, but plugging values in to check the answers and graphing them revealed a graph that[/q1]
[q1]> was similar, but slightly off. If anyone has any ideas, PLEASE let me know, I could really use[/q1]
[q1]> some extra credit![/q1]
[q1]>[/q1]
[q1]> Thanks, -Kelley[/q1]
[q1]>[/q1]
[q1]> PS, if anyone wants me to, I can send them an attatchment of the problem that has the correct[/q1]
[q1]> symbols that I can't post here.[/q1]
Well, lovely maple-whic-cannot-be-wrong (god Uni makes you lazy) says that the integral is 1/2
ln (ln x^2).
Robbo wrote:
[q1]> What is "u prime?" I have no idea, I have never heard of the way you did it. Only your[/q1]
[q1]> instructor's way. I only did Methods, perhaps an HL student would like to explain. Hmmm... my[/q1]
[q1]> maths is slowly going out the window I think...[/q1]
[q1]>[/q1]
[q1]> Cidra17918 wrote:[/q1]
[q1]>[/q1]
[q2]> > This was a problem my SL calculus class did for a warm-up. I used one method, my instructor used[/q2]
[q2]> > another, both were correct, and the answers were different. Now it's extra credit to figure out[/q2]
[q2]> > why the answers were different, and we're allowed to seek outside help.[/q2]
[q2]> >[/q2]
[q2]> > Since the symbol for integration can't be used on this message board, I'm going to use S[/q2]
[q2]> > instead. It will probably help if you write this out on a piece of paper using the real symbols[/q2]
[q2]> > for integrate, absolute value, and squared.[/q2]
[q2]> >[/q2]
[q2]> > The problem is as follows:[/q2]
[q2]> >[/q2]
[q2]> > S (1) / (x ln (x squared)) dx[/q2]
[q2]> >[/q2]
[q2]> > MY METHOD[/q2]
[q2]> >[/q2]
[q2]> > S (1) / (x ln (x squared)) dx[/q2]
[q2]> >[/q2]
[q2]> > S (1/x) / (ln x squared) dx[/q2]
[q2]> >[/q2]
[q2]> > dx= (1/2) S 2 (absolute value of x) / ln x squared dx[/q2]
[q2]> >[/q2]
[q2]> > u= ln x squared[/q2]
[q2]> >[/q2]
[q2]> > u prime= 2/x dx[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Is u prime the derivative of u?[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > = (1/2) ln (absolute value of u) + C[/q2]
[q2]> >[/q2]
[q2]> > =(1/2) ln (absolute value of ln x squared) + C[/q2]
[q2]> >[/q2]
[q2]> > = (1/2) ln (absolute value of 2 ln x) + C[/q2]
[q2]> >[/q2]
[q2]> > = ln (absolute value of ln x) + C[/q2]
[q2]> >[/q2]
[q2]> > MY INSTRUCTORS METHOD[/q2]
[q2]> >[/q2]
[q2]> > S (1) / (x ln (x squared)) dx[/q2]
[q2]> >[/q2]
[q2]> > = S (1) / (2x ln x) dx[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> This line above uses the fact that you take the x squared on the ln x and put it in front as a[/q1]
[q1]> factor of the x.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > = (1/2) S (1) / (x ln x) dx[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Take the two out the front.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > =(1/2) (1/x) (1/ ln x) dx=du[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Put each term separately.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > u= ln x[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Substitution method.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > du= (1/x) dx[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> More substitution method.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > = (1/2) S (1/u) du[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Back to original equation using susbtituted values.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > = (1/2) ln (absolute value of u) + C[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Now back to x, substituting back from u.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > = (1/2) ln (absolute value of ln x) + C[/q2]
[q2]> >[/q2]
[q2]> > Now we're posed with the problem of finding out what went wrong. As far as I can see, both[/q2]
[q2]> > methods were correct, but plugging values in to check the answers and graphing them revealed a[/q2]
[q2]> > graph that was similar, but slightly off. If anyone has any ideas, PLEASE let me know, I could[/q2]
[q2]> > really use some extra credit![/q2]
[q2]> >[/q2]
[q2]> > Thanks, -Kelley[/q2]
[q2]> >[/q2]
[q2]> > PS, if anyone wants me to, I can send them an attatchment of the problem that has the correct[/q2]
[q2]> > symbols that I can't post here.[/q2]
ln (ln x^2).
Robbo wrote:
[q1]> What is "u prime?" I have no idea, I have never heard of the way you did it. Only your[/q1]
[q1]> instructor's way. I only did Methods, perhaps an HL student would like to explain. Hmmm... my[/q1]
[q1]> maths is slowly going out the window I think...[/q1]
[q1]>[/q1]
[q1]> Cidra17918 wrote:[/q1]
[q1]>[/q1]
[q2]> > This was a problem my SL calculus class did for a warm-up. I used one method, my instructor used[/q2]
[q2]> > another, both were correct, and the answers were different. Now it's extra credit to figure out[/q2]
[q2]> > why the answers were different, and we're allowed to seek outside help.[/q2]
[q2]> >[/q2]
[q2]> > Since the symbol for integration can't be used on this message board, I'm going to use S[/q2]
[q2]> > instead. It will probably help if you write this out on a piece of paper using the real symbols[/q2]
[q2]> > for integrate, absolute value, and squared.[/q2]
[q2]> >[/q2]
[q2]> > The problem is as follows:[/q2]
[q2]> >[/q2]
[q2]> > S (1) / (x ln (x squared)) dx[/q2]
[q2]> >[/q2]
[q2]> > MY METHOD[/q2]
[q2]> >[/q2]
[q2]> > S (1) / (x ln (x squared)) dx[/q2]
[q2]> >[/q2]
[q2]> > S (1/x) / (ln x squared) dx[/q2]
[q2]> >[/q2]
[q2]> > dx= (1/2) S 2 (absolute value of x) / ln x squared dx[/q2]
[q2]> >[/q2]
[q2]> > u= ln x squared[/q2]
[q2]> >[/q2]
[q2]> > u prime= 2/x dx[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Is u prime the derivative of u?[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > = (1/2) ln (absolute value of u) + C[/q2]
[q2]> >[/q2]
[q2]> > =(1/2) ln (absolute value of ln x squared) + C[/q2]
[q2]> >[/q2]
[q2]> > = (1/2) ln (absolute value of 2 ln x) + C[/q2]
[q2]> >[/q2]
[q2]> > = ln (absolute value of ln x) + C[/q2]
[q2]> >[/q2]
[q2]> > MY INSTRUCTORS METHOD[/q2]
[q2]> >[/q2]
[q2]> > S (1) / (x ln (x squared)) dx[/q2]
[q2]> >[/q2]
[q2]> > = S (1) / (2x ln x) dx[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> This line above uses the fact that you take the x squared on the ln x and put it in front as a[/q1]
[q1]> factor of the x.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > = (1/2) S (1) / (x ln x) dx[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Take the two out the front.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > =(1/2) (1/x) (1/ ln x) dx=du[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Put each term separately.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > u= ln x[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Substitution method.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > du= (1/x) dx[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> More substitution method.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > = (1/2) S (1/u) du[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Back to original equation using susbtituted values.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > = (1/2) ln (absolute value of u) + C[/q2]
[q2]> >[/q2]
[q1]>[/q1]
[q1]> Now back to x, substituting back from u.[/q1]
[q1]>[/q1]
[q2]> >[/q2]
[q2]> > = (1/2) ln (absolute value of ln x) + C[/q2]
[q2]> >[/q2]
[q2]> > Now we're posed with the problem of finding out what went wrong. As far as I can see, both[/q2]
[q2]> > methods were correct, but plugging values in to check the answers and graphing them revealed a[/q2]
[q2]> > graph that was similar, but slightly off. If anyone has any ideas, PLEASE let me know, I could[/q2]
[q2]> > really use some extra credit![/q2]
[q2]> >[/q2]
[q2]> > Thanks, -Kelley[/q2]
[q2]> >[/q2]
[q2]> > PS, if anyone wants me to, I can send them an attatchment of the problem that has the correct[/q2]
[q2]> > symbols that I can't post here.[/q2]
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