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Question Statement

My attempt:

Attempt

Relooking at it, I think $L_{20}$ would be the highest, so like

$R_{1200} < L_{1200} < L_{20}$, but I have no way to justify it, any help is appreciated.

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    This is a good attempt, just try and work on [typesetting](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) and things'll be looking really good. – 2017-01-17
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    The fact that $f$ is decreasing is a good start. Now draw some pictures. To simpify, first think about say $L_{20}$ and $L_{40}$, then think about $L_{1200}$ later. – 2017-01-17
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    Draw the curve and L10 and R10.You will see the answer. – 2017-01-17

3 Answers 3

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No matter what $n$ and $m$ are, $R_n

Now both $L_{20}$ and $L_{1200}$ overestimate the value of $A$. Informally, $L_{1200}$ is closer to $A$, because $A=\lim_{n\to\infty}L_{n}$. The function is not particularly weird enough for $L_{1200}$ to break the downward trend of $L_n$ towards $A$ as $n\to\infty$. So this much understanding suggests $L_{1200}$ is the smaller of the remaining two numbers.

A little more formally, $L_{20}$ is the area of a certain $20$ rectangles. And $L_{1200}$ is the area of a certain $1200$ rectangles. Since $20$ divides into $1200$, we can in fact place sets of $60$ of the rectangles from $L_{1200}$ inside each rectangle from $L_20$. Since the function is deacreasing, the $60$ rectangles will fit inside the one rectangle with room to spare. So again, $L_{1200}$ should be less than $L_{20}$.

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The main ideas here are that, for a decreasing function,

  • Estimating using left endpoints (i.e., $L_{20}, L_{1200}$) will always overestimate the true value of the integral.

  • Estimating using right endpoints will always underestimate the value of the integral.

These two tell you that $R_m < I < L_n$ for any integers $m, n$.

  • Using more subintervals (i.e., larger $n$) will always produces a better estimate -- they'll produce estimates closer to the actual value of the integral.

So, if $m < n$, we have

$$R_m < \underbrace{R_n < I < L_n}_{\text{more subintervals}} < L_m$$

which agrees exactly with what you have! (in the body of the question, not the image)

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You're last thought is correct. One rigorous justification goes as follows:

$\displaystyle L_{20} = \sum_{i=0}^{19}\frac{5}{20}f\left(9+\frac{5i}{20}\right)$, and

$\displaystyle L_{1200} = \sum_{i=0}^{1199}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right)$.

Now, we are going to split up the $L_{1200}$ sum into groups of 60. In particular,

$\displaystyle L_{1200} = \sum_{i=0}^{59}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right) + \sum_{i=60}^{119}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right) + \ldots + \sum_{i=1140}^{1199}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right)$.

Now, by monotonicity of the function $f$ we have that this sum is less than

$\displaystyle\sum_{i=0}^{59}\frac{5}{1200}f\left(9\right) + \sum_{i=60}^{119}\frac{5}{1200}f\left(9+\frac{5\cdot 60}{1200}\right) + \ldots + \sum_{i=1140}^{1199}\frac{5}{1200}f\left(9+\frac{5\cdot 1140}{1200}\right) \\ =60\cdot\frac{5}{1200}f\left(9\right) + 60\cdot\frac{5}{1200}f\left(9 + \frac{5}{20}\right) + \ldots + 60\cdot\frac{5}{1200}f\left(9 + \frac{5\cdot 19}{20}\right) =: L_{20}$.

Therefore, $L_{1200} \leq L_{20}$.