Help with inequality confusing inequality question (1 Viewer)

willythefree · Feb 7, 2015

Please help, its so confusing, I don't know where to begin in regards to part c).
Thanks to anyone that helps

InteGrand · Feb 7, 2015

willythefree said:
Please help, its so confusing, I don't know where to begin
Thanks to anyone that helps View attachment 31773

Here's how to do part a).

Consider the area A_j of the rectangle between x = j and x = j+1, where $j\in \left \{ \left. 2,3,4,\cdots,k \right \} \right.$

It's width is $(j+1)-1=1$ and its height is $\ln j$ . Hence its area is $A_j = 1\cdot\ln j = \ln j$ .

Since j varies over the integers from 2 to k (since we're told the rectangles are between x = 2 and x = k+1, so the last rectangle has its left side at x = k), the total area of the rectangles is

$A_{\text{tot.}}=\sum_{j=2}^{k}A_j$

$=\sum_{j=2}^{k}\ln j$

$=\ln 2 + \ln 3 + \ln 4 + \cdots + \ln k = S$ .

integral95 · Feb 7, 2015

you can see from the diagram that

$\int_{2}^{k+1}ln(x-1)\ \ dx\ < S < \int_{2}^{k+1}ln(x)\ \ dx$

From there you manipulate the inequality to get your solution
(which isn't that easy if you're not too familiar with your log laws and exponential)

Sy123 · Feb 7, 2015

http://i.imgur.com/9CYSHXz.png

Help with inequality confusing inequality question (1 Viewer)

willythefree

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InteGrand

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integral95

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Sy123

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