Definite Integral and ln

[Jason76]

\(\displaystyle \int^{10}_{5}\dfrac{\ln(10x)}{x}dx\) - Answer 2.592

Any clue on where to start? Integration by Parts?

 

[MarkFL]

I would write:

\(\displaystyle \displaystyle I=\int_5^{10}\frac{\ln(10x)}{x}\,dx=\int_5^{10} \frac{\ln(10)+\ln(x)}{x}\,dx=\ln(10)\int_5^{10} \frac{1}{x}\,dx+\int_5^{10} \frac{\ln(x)}{x}\,dx\)

The first integral is straightforward, and the second requires only a \(\displaystyle u\)-substitution:

\(\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx\)

\(\displaystyle \displaystyle I=\ln(10)(\ln(10-\ln(5))+\int_{\ln(5)}^{\ln(10)}u\,du \)

Can you proceed?

edit: You should find the result is approximated by 2.952...

 

[MarkFL]

Upon further reflection, we need not split the numerator of the integrand at all:

\(\displaystyle \displaystyle \int_5^{10}\frac{\ln(10x)}{x}\,dx\)

Use the substitution:

\(\displaystyle \displaystyle u=\ln(10x)\,\therefore\,du=\frac{1}{x}\,dx\)

and we have:

\(\displaystyle \displaystyle \int_{\ln(50)}^{\ln(100)}u\,du\)

Now the same result is more easily obtained.

 

[Jason76]

I'm wondering if terms could be rearranged so to use "integration by Parts" with \(\displaystyle u = \ln(10)\) and \(\displaystyle v = 1\) But probably not.

 

[srmichael]

I'm wondering if terms could be rearranged so to use "integration by Parts" with \(\displaystyle u = \ln(10)\) and \(\displaystyle v = 1\) But probably not.

Why would you wanna do that when you don't have to?

 

[Jason76]

\(\displaystyle \displaystyle I=\int_5^{10}\frac{\ln(10x)}{x}\,dx=\int_5^{10} \frac{\ln(10)+\ln(x)}{x}\,dx=\ln(10)\int_5^{10} \frac{1}{x}\,dx+\int_5^{10} \frac{\ln(x)}{x}\,dx\)

Got it up to this point.

 

[MarkFL]

I'm wondering if terms could be rearranged so to use "integration by Parts" with \(\displaystyle u = \ln(10)\) and \(\displaystyle v = 1\) But probably not.

Yes, you could use integration by parts here, but you should learn how to correctly use IBP first. Essentially, we use:

\(\displaystyle \displaystyle \int u\,dv=uv-\int v\,du\)

So, you want to choose a suitable \(\displaystyle u\) and \(\displaystyle dv\) from the original integral. Using the LIATE rule, we could use:

\(\displaystyle u=\ln(10x)\,\therefore\,du=\frac{1}{x}dx\)

\(\displaystyle dv=\frac{1}{x}\,dx\,\therefore\,v=\ln(x)\)

and we have:

\(\displaystyle \displaystyle \int_5^{10}\frac{\ln(10x)}{x}\,dx=\left[\ln(10x)\ln(x) \right]_5^{10}-\int_5^{10}\frac{\ln(x)}{x}dx\)

\(\displaystyle \displaystyle \int_5^{10}\frac{\ln(10x)}{x}\,dx=\ln(100)\ln(10)-\ln(50)\ln(5)-\int_5^{10}\frac{\ln(x)}{x}dx\)

Can you proceed? You can either use a substitution or IBP again on the remaining integral.

Using this method for practice is fine, but most people seek to use the simplest and most straightforward technique possible. I recommend first looking to see if the integral can be integrated directly, if not, then look for a substitution that will transform the integral into a form that can, and if not, then look to see if IBP can be used. It is good to know as many ways as possible, that way if you miss a substitution that will work, you have other techniques to fall back on.

 

[MarkFL]

\(\displaystyle \displaystyle I=\int_5^{10}\frac{\ln(10x)}{x}\,dx=\int_5^{10} \frac{\ln(10)+\ln(x)}{x}\,dx=\ln(10)\int_5^{10} \frac{1}{x}\,dx+\int_5^{10} \frac{\ln(x)}{x}\,dx\)

Got it up to this point.

I would work the problem out using the second method I posted first...it is the simplest of those I have posted.

 

[Jason76]

Yes, you could use integration by parts here, but you should learn how to correctly use IBP first. Essentially, we use:

\(\displaystyle \displaystyle \int u\,dv=uv-\int v\,du\)

So, you want to choose a suitable \(\displaystyle u\) and \(\displaystyle dv\) from the original integral. Using the LIATE rule, we could use:

\(\displaystyle u=\ln(10x)\,\therefore\,du=\frac{1}{x}dx\)

\(\displaystyle dv=\frac{1}{x}\,dx\,\therefore\,v=\ln(x)\)

and we have:

\(\displaystyle \displaystyle \int_5^{10}\frac{\ln(10x)}{x}\,dx=\left[\ln(10x)\ln(x) \right]_5^{10}-\int_5^{10}\frac{\ln(x)}{x}dx\)

\(\displaystyle \displaystyle \int_5^{10}\frac{\ln(10x)}{x}\,dx=\ln(100)\ln(10)-\ln(50)\ln(5)-\int_5^{10}\frac{\ln(x)}{x}dx\)

Can you proceed? You can either use a substitution or IBP again on the remaining integral.

Using this method for practice is fine, but most people seek to use the simplest and most straightforward technique possible. I recommend first looking to see if the integral can be integrated directly, if not, then look for a substitution that will transform the integral into a form that can, and if not, then look to see if IBP can be used. It is good to know as many ways as possible, that way if you miss a substitution that will work, you have other techniques to fall back on.

Because of limited time, would have to use this method cause more familiar with IBP.

 

[MarkFL]

If you're looking to save time, then it would behoove you to learn simpler methods. IBP is one of the costlier methods when it comes to time.

I have never heard of anyone learning IBP before substitutions.