Natural logarithmic function problem

[Xcron]

This is the problem:

(A) By comparing areas, show that
\frac{1}{3}\ll\ln1.5\ll\frac{5}{12}

(B) Use the Midpoint Rule with n=10 to esimate \ln1.5.I've seen these types of "comparing areas" problems but I kind of forgot how to go about solving the problem...I was thinking perhaps using e^x on the three parts but that was a plausible solution to this, really...the midpoint part is also somewhat strange because all three are constants so...there is really an f(x) type of function to work with for the Riemann sum type of set-up...

Any help/advice?

 

[benorin]

Recall that

\ln a=\int_{x=1}^{a} \frac{1}{x}dx

and since \frac{1}{x} is an increasing function of x, one may say that the above integral is bounded above by the sum using righthand end-points and below by the sum taken with lefthand end-points. end-points.

 

[d_leet]

Recall that
\ln a=\int_{x=1}^{a} \frac{1}{x}dx
and since \frac{1}{x} is an increasing function of x, one may say that the above integral is bounded above by the sum using righthand end-points and below by the sum taken with lefthand end-points. end-points.

Wouldn't that be the other way around because \frac{1}{x} is strictly decreasing for x > 0. So the left hand end point sums should be greater than those taken at the right hand.

 

[benorin]

Yep, I was thinking of ln(x) being increasing, sorry: my bad.

 

[Xcron]

I'm trying to figure out how to use the Midpoint Rule for part (B) and I seem to be getting the wrong answer. The point of the rule is to use midpoints with 10 sub-intervals...so I thought I could just use:
\frac{1}{10}[f(\frac{40.5}{120})+f(\frac{41.5}{120})+...+f(\frac{49.5}{120})]

But that doesn't seem to be working...hmmm...The next problem is very similar and I am somewhat confused with it as well...the problem says:

By comparing areas, show that
\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\ll\ln n\ll 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}

I thought I could uise a riemann sum for the comparison since I fugred if I could show that the very left part is less than the very right (because the right part has a 1 added to it at the very beginning). However, I'm not sure how to show that \ln n is between those two...

 

[krab]

Why are you using the double inequality sign? This means MUCH less than, or MUCH greater than. Just use the symbols directly (Shift-, or Shift-. on American keyboards).

 

[Xcron]

Can anyone help me with the Midpoint Rule problem...and the new Riemann sum problem please?

 

[Xcron]

I think that for the Midpoint Rule problem...I would have sub-intervals(10) represented by \frac{1}{10}, which would be in front of the Riemann sum. I probably should not use the f(\frac{40.5}{120})...because they seem a bit strange...I'm trying to figure out how to estimate ln 1.5 while doing a sum of multiple values...*sigh*..

For the second "comparing areas" problem...I think I might be able to set up:
\int_{0}^{1} \frac{1}{x}dx - 1 &lt; \int_{0}^{1} \frac{1}{x}dx &lt; <br /> \int_{0}^{1} \frac{1}{x}dx - ?? ... I'm not sure...