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April 27th, 2018, 03:25 PM   #1
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Upper darboux integral x^3

I'm trying to calculate the upper darboux integral of $f(x) = x^3$ on $[0, b]$ where $b$ is a positive integer.

Let $P$ be a partition defined as $\lbrace x_0 = 0, x_1, x_2, ... ,x_n = b \rbrace$, $x_i < x_{i+1}$.

Define $M(f; S) = \sup\lbrace x^3 : x \epsilon S \rbrace$

Define Upper Darboux sum as $U(f; P) = \sum_{i = 1}^{n} (M(x^3 ; [x_{i-1}, x_i]) \cdot (x_i - x_{i-1}))$

So $U(f; P) = \sum_{i = 1}^n x_i^3 \cdot(x_i - x_{i-1})$

In the example for $f(x) = x^2$, we have to choose some $x_i$ to make the algebra pan out... but why are we choosing an $x_i$ ? I don't get why the next step is to choose $x_i = ...$

Please no full solution, thank you

Last edited by Choboy11; April 27th, 2018 at 03:28 PM.
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April 28th, 2018, 12:05 PM   #2
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For $f(x)=x^2$, the approach is the same as for $f(x)=x^3$, replacing $x_i^3\ by \ x_i^2$ in the expression for U. I don't understand your last remark.
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April 28th, 2018, 01:34 PM   #3
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Sorry fro the confusion

The full question I am working on is "Find the upper and lower Darboux integrals for $f(x) = x^3$ on the interval $[0, b]$."


$U(f, P) = \sum_{i=1}^n M(f(x), [x_{i-1}, x_i]) \cdot \triangle x$
$= \sum_{i=1}^n x_i^3(x_i - x_{i-1})$

Similar to the example in my book, let $x_i = i \cdot \frac bn$. I realize $\frac bn$ is the value of $x_i - x_{i-1}$ assuming the $x_i$'s are spaced out equally, but why do we do this step? Why are we specifying a partition(i think thats what we are doing here...)?

After that I get,

$U(f,P) = \sum_{i=1}^n i^3 (\frac bn) ^3 \cdot \frac bn$
$= (\frac bn)^4 \cdot (\sum_{i=1}^n i)^2$
$= (\frac bn)^4 \cdot (\frac{n(n+1)}{2})^2$
$= (\frac bn)^4 \cdot (\frac {n^4 + 2n^3 + n^2}{4}$.

So $U(f,P) \rightarrow \frac {b^4}{4}$ as $n \rightarrow \infty$.

By similar steps, we can show $L(f, P) = (\frac bn)^4 \cdot \frac{n^4 - 2n^3 +n^2}{4} \rightarrow \frac {b^4}{4}$ as $n \rightarrow \infty$.

We can also conclude $U(f, P) > L(f, P) \ge \frac{b^4}{4}$. Here I am stuck.

In the book example (for f(x) = x^2), they have $U(f,P) \le (\frac {b^3}{3})$ and $L(f,P) \ge (\frac {b^3}{3}$ and conclude that f is integrable. I don't understand this logic.

As i understand it, to show $f$ is integrable on $[a, b]$, we need to show
$\inf \lbrace U(f, P) : \text{P is a partition of }[a,b]) \rbrace = \sup \lbrace L(f, P) : \text{P is a partition of [a,b]}) \rbrace$.

Last edited by Choboy11; April 28th, 2018 at 01:43 PM.
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April 28th, 2018, 02:43 PM   #4
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If I understand you correctly, you are asking exactly the right question. In general, you have to show this limit exists for any partition. However, this is what makes Darboux sums so nice to work with. There is a theorem which says Darboux integrability is equivalent to Riemann integrability. With this in hand, you are free to choose a partition. The reason this works is sketched below.

1. Suppose you pick a sequence of partitions, $P_n$ whose mesh size goes to $0$, and you show that $U(f,P_n) - L(f,P_n) \to 0$.

2. Now, prove that if $P$ is a partition and $Q$ is any refinement of $P$ (i.e $P \subset Q$), then $L(f,P) \leq L(f,Q) \leq U(f,Q) \leq U(f,P)$.

3. Now, it follows that if $Q_n$ is any other sequence of partitions whose mesh goes to $0$, then $Q_n \cup P_n$ is a common refinement. Combine this with (1),(2), and the squeeze theorem to conclude the result.
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May 1st, 2018, 09:46 PM   #5
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Sorry for the late reply... finals are coming and we started power series so i've put integration on the back burner... for now.

So far i've proved part 2(read it in my book... but this is it in my own words i guess?)

Suppose $P, Q$ are partitions of some interval $[a, b]$ such that $P \subset Q$ and $P = \lbrace x_0 < x_1 < ... < x_n\rbrace$ and $Q = \lbrace x_0 < x_1 < ... < x_{k-1} < x_k < x_{k+1} < ...x_n \rbrace$. (EDIT: I still need to explain why choosing these partitions is sufficient to show the statement holds for all $P, Q$ such that $P \subset Q$.)

Observe $$U(f, P) - U(f, Q) = M(f, [x_{k-1}, x_{k+1}])\cdot (x_{k+1} - x_{k-1}) - M(f, [x_{k-1}, x_k]) \cdot (x_k - x_{k-1}) - M(f, [x_k, x_{k+1}])\cdot(x_{k+1} - x_k) \ge 0$$. (The $\ge 0$ step seems intuitive, do I need to justify it?)

Thus we can conclude $U(f, P) - U(f, Q) \ge 0$ so $U(f, P \ge U(f, Q)$. By similar logic $L(f, P) \le L(f, Q)$. It is obvious.. i guess.. that $L(f, Q) \le U(f, Q)$. Combining these statements gives us $L(f, P) \le L(f, Q) \le U(f, Q) \le U(f, P)$.

I had a few questions in the proof about if I need further justification.

I'll post again when I get 1, 3. Thank you for your time

Last edited by Choboy11; May 1st, 2018 at 09:55 PM.
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May 14th, 2018, 10:52 PM   #6
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Since f(x) is continuous, the Darboux integral is the Riemann integral by definition.*

But I think the point here is to show the actual mechanics of the definition because in general f doesn't have to be continuous.

If P= {1,2,3), by definition

$\displaystyle U(f,P)= 2^{3}(2-1) + 3^{3}(3-2)$

The Upper Darboux integral is the infimum of U(f,P) for all P, not just equal intervals. That sounds impossible.

For equal intervals, you can write an expression for the sum for any division
and you have to show the sum is a minimum as number of subdivisions approaches $\displaystyle \infty$.

Darboux Integral -- from Wolfram MathWorld

Interesting. I always wondered what a Darboux integral was.

Actually, the Riemann integral is based on the max interval approaching zero, and it's not obvious why that result should be the same as the Darboux integral, for continuous functions, but I believe you can prove RI is the same for all equal intervals approahing 0 and max interval approaching zero,

Last edited by zylo; May 14th, 2018 at 11:11 PM.
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