Need help on step by step solution for this equation

[hahamonkey]

[math]\int_{y}^{\ell}\frac{1}{\ell x} dx = \frac{1}{\ell} log \frac{\ell}{y}[/math]
I got this equation from watching a professor's video on solving continuous probability.

I'm bad at calculus, algebra, and logarithms. I don't understand how the RHS is derived from the LHS.
I have tried plugging LHS into wolfram but it failed to produce a solution. Also tried Microsoft math solver but it gave out a different answer: [imath]\frac{ln(|\ell|)-ln(|y|)}{\ell}[/imath]

I would really appreciate a step by step solution and how it's derived, thank you.

 

[blamocur]

If you are bad in all those subjects it might be really difficult to understand this derivation.
BTW, the answer from Microsoft is equivalent to the one in your professor's video, at least for positive [imath]l[/imath] and [imath]y[/imath].

 

[Harry_the_cat]

Do you see that \(\displaystyle \int_{y}^{l}\frac{1}{lx}\,dx\) = \(\displaystyle \frac{1}{l}\int_{y}^{l}\frac{1}{x}\,dx\) ?

Can you finish it from there?

 

[Deleted member 4993]

I'm bad at calculus, algebra, and logarithms.

Did you take those courses for credit? Did you pass?

 

[Steven G]

\(\displaystyle \int \dfrac {1}{x}dx= \ln x + c\)

 

[Steven G]

BTW \(\displaystyle \ln a - \ln b = \ln \dfrac {a}{b}\)

 

[lookagain]

\(\displaystyle \int \dfrac {1}{x}dx= \ln x + c\)

\(\displaystyle \int \dfrac {1}{x}\ dx \ = \ ln|x| \ + \ C\)

 

[Steven G]

\(\displaystyle \int \dfrac {1}{x}\ dx \ = \ ln|x| \ + \ C\)

I know. I chose not to put the absolute value bars because the solution didn't have any and I my goal was for the OP to see what was going on.

 

[hahamonkey]

[imath]\int_y^\ell \frac{1}{\ell x} dx \\= \frac{1}{\ell} \int_y^\ell \frac{1}{x}dx \\= \frac{1}{\ell}(ln(\ell)-ln(y)) \\= \frac{1}{\ell}ln(\frac{\ell}{y})[/imath]

Ok I think I got it. The hints are really helpful.

Thanks @Harry_the_cat, @lookagain, @Steven G. You guys are great.

 

[lookagain]

I know. I chose not to put the absolute value bars because the solution didn't have any and I my goal was for the OP to see what was going on.

The portion of the work from the professor the student bothered showing us does not have the absolute bars, but they are part of the work. What you show for a student is to be a correct step regardless of what possible shortcomings of professors' knowledge or missing intermediate steps.
Then, if the quantities l (el) and y are necessarily
positive for this problem, because it has to do with continuous probabilities, you can mention
that the absolute value bars will later be dropped.

 

[Steven G]

[imath]\int_y^\ell \frac{1}{\ell x} dx \\= \frac{1}{\ell} \int_y^\ell \frac{1}{x}dx \\= \frac{1}{\ell}(ln(\ell)-ln(y)) \\= \frac{1}{\ell}ln(\frac{\ell}{y})[/imath]

Ok I think I got it. The hints are really helpful.

Thanks @Harry_the_cat, @lookagain, @Steven G. You guys are great.

Now I will point out that your professor has a mistake.
\(\displaystyle \int \dfrac{1}{x}dx = \ln |x|\). Your professor left out the absolute value bars.
Can you please update you answer.