Area bounded by f(x) and x-axis

[Guest]

Hi guys, I'm having a tough week and so I could just be overlooking something simple because I am kaput. Anyway... My problem is:

Let ƒ(x) = log(x) ÷ x, x in the interval [1, 2e]

Part a) Find the area of the region R bounded by the graph of ƒ and the x-axis.

Part b) Find the volume of the solid generated by revolving R about the x-axis.

So I figured for a I have to subtract the two integrals, right? So I tried to set it up like...

∫ (log(x) ÷ x) - ∫ 0

And I set the integrals from 1 to 2e. But if I integrate log(x) ÷ x I would get a quotient rule return of:
((1÷x)(x)-(log x)(1))/x^2

(Sorry, easier to type ^2 on a Mac)

Again, from 1 to 2e. Correct? So then plugging in the values I get:

(1 - (log 2e/2e^2)) - (1 - log 1)

Right? I must have done something wrong... That seems like a fairly strange answer.

Thanks in advance for any help!

 

[pka]

\(\displaystyle \int {\frac{{\ln \left( x \right)}}{x}dx = \frac{1}{2}\ln ^2 (x)}\)

 

[Guest]

Oh, awesome, thanks!

So then evaluating that as my assignment gives 2, correct?

My friend just told me that I don't need the - 0 in the original equation... Is that valid? The area is just 2 units?

 

[galactus]

part b:

\(\displaystyle \L\\{\pi}\int_{1}^{2e}\left(\frac{ln(x)}{x}\right)^{2}dx\)

 

[Guest]

Excellent, thanks for all the help! I think I get it now!

 

[tkhunny]

Let's work on your language just a bit.

I set the integrals from 1 to 2e.

You what? Do you mean the Limits of Integration? How did you find those values?

If I integrate log(x) ÷ x I would get a quotient rule return of: ((1÷x)(x)-(log x)(1))/x^2

What? Why are you finding the derivative when you should be finding the ANTI-derivative?