Proving x^2 + x is integrable.

1. Dec 10, 2012

Zondrina

1. The problem statement, all variables and given/known data

I'm curious about how to do something like this. My professor gave us a very simple example of how to prove f(x) = k is integrable on [1,3] for k in ℝ.

I'm trying to apply the same logic to any function so I can see how it works. So I guess I could state the problem as :

Suppose $f(x) = x^2 + x$, prove f is integrable on [2,5] by showing I = sup{sp} = inf{Sp} = J.

2. Relevant equations

sp is my underestimate and Sp is my overestimate.

3. The attempt at a solution

So first off we note :

mi = inf{ f(x) | x in [2,5] } = 6
Mi = sup{ f(x) | x in [2,5] } = 30

Forming our upper and lower sums and then calculating them :

$s_p = \sum_{i=1}^{n} m_i Δx_i = 6(3) = 18$

$S_p = \sum_{i=1}^{n} M_i Δx_i = 30(3) = 90$

This is definitely not what's supposed to happen I think as we all intuitively know that f(x) is integrable in this case. Showing it is a bit difficult for me though. Could someone point out what I may have done wrong?

2. Dec 10, 2012

Ray Vickson

These are NOT what are meant by S_p and s_p. You need to use
$$m_i = \inf_{x_i \leq x \leq x_i + \Delta x_i} f(x) ,\;\;\;\;\; M_i = \sup_{x_i \leq x \leq x_i + \Delta x_i} f(x), \; i = 1,2 \ldots, n .$$

3. Dec 10, 2012

pasmith

Let $P$ be a partition of [2,5] with points of division $2 = x_0 < x_1 < \dots < x_{n-1} < x_n = 5$, and let $\delta_i = x_i - x_{i-1}$ for $1 \leq i \leq n$. Then you should have
$$m_i = \inf \{x^2 + x : x \in [x_{i-1},x_i] \} = x_{i-1}^2 + x_{i-1} \\ M_i = \sup \{x^2 + x : x \in [x_{i-1},x_i] \} = x_i^2 + x_i$$
since $x^2 + x$ is strictly increasing on [2,5].

Now
$$x_{i-1}(x_i-x_{i-1}) < \frac12(x_i+x_{i-1})(x_i-x_{i-1}) = \frac12 (x_i^2 - x_{i-1}^2) \\ x_{i-1}^2(x_i- x_{i-1}) < \frac13(x_i^2 + x_i x_{i-1} + x_{i-1}^2)(x_i - x_{i-1}) = \frac13 (x_i^3 - x_{i-1}^3)$$
and in the other direction
$$x_{i}(x_i-x_{i-1}) > \frac12(x_i+x_{i-1})(x_i-x_{i-1}) = \frac12 (x_i^2 - x_{i-1}^2) \\ x_{i}^2(x_i- x_{i-1}) > \frac13(x_i^2 + x_i x_{i-1} + x_{i-1}^2)(x_i - x_{i-1}) = \frac13 (x_i^3 - x_{i-1}^3)$$
so
$$s(P) = \sum m_i \delta_i = \sum x_{i-1}^2(x_i- x_{i-1}) + \sum x_{i-1}(x_i-x_{i-1}) < \sum \frac13 (x_i^3 - x_{i-1}^3) + \sum \frac12 (x_i^2 - x_{i-1}^2)\\ = \frac13(5^3 - 2^3) + \frac12 (5^2 - 2^2)$$
by telescoping and similarly $S(P) = \sum M_i \delta_i > \frac13(5^3 - 2^3) + \frac12 (5^2 - 2^2)$.

Since these limits are independent of $\max \delta_i$, we must have
$$\sup s(P) \leq \frac13(5^3 - 2^3) + \frac12 (5^2 - 2^2) \leq \inf S(P)$$
Now one just has to prove equality.

It's slightly easier to prove for the Riemann integral: $x^2 + x$ is continuous, so by the intermediate value theorem there exists $z_i \in [x_i,x_{i-1}]$ such that
$$z_i^2 + z_i = \frac13(x_i^2 + x_ix_{i-1} + x_{i-1}^2) + \frac12(x_i + x_{i-1})$$.
Then
$$\sum_{i} (z_i^2 + z_i)(x_i - x_{r-1}) = \sum_{i} \frac13(x_i^3-x_{i-1}^3) + \frac12 (x_i^2 - x_{i-1}^2) = \frac13(5^3 - 2^3) + \frac12(5^2 - 2^2)$$
and since the result is independent of $\max \delta_i$ it must be the limit as $\max \delta_i \to 0$.

Of course I'd never have come up with those inequalities if I didn't already know the answer.

4. Dec 10, 2012

Zondrina

Ahhh that was the trick. I completely forgot to partition my interval into n equal sub intervals ( Derp moment ).

Then $x_i = 5i/n$ and $x_{i-1} = 5(i-1)/n$ so that $m_i = x_{i-1} = 5(i-1)/n$ and $M_i = x_i = 5i/n$

Thus :

$S_p = \sum_{i=1}^{n} M_i Δx_i = 25/n^2 \sum_{i=1}^{n} i$

( We do a similar case for the lower sum ).

After evaluating and simplifying that, we get J = inf{ Sp } ≤ Sp and since inf{ Sp } does not rely on n, we can observe what happens as n→∞ which will yield our result for J. We can do the same thing for I and the lower sum.

Since I = J, we deduce that f is integrable on [2,5] and its common value is 99/2.

EDIT : Hmmm after writing this out I get I = 25/2 = J which is sorta what I wanted to happen, except I want 99/2, not 25/2.

Did i partition incorrectly?

Last edited: Dec 10, 2012
5. Dec 10, 2012

pasmith

I'll just do this for $f(x) = x$ on $[a,b]$ with $a \geq 0$; the same idea should work for $f(x) = x^2$.

For the upper sum, I want to show that for all $\epsilon > 0$ there exists a paritition $P$ such that
$$\frac 12(b^2 - a^2) < S(P) < \frac 12(b^2 - a^2) + \epsilon$$
So I have
$$S(P) = \sum_i x_i (x_i - x_{i-1}) = \sum_i \frac{x_i + x_{i-1} + x_i - x_{i-1}}{2}(x_i - x_{i-1}) = \sum_i \frac12 (x_i^2 - x_{i-1}^2) + \sum_i \frac12 (x_i - x_{i-1})^2 \\ = \frac12 (b^2 - a^2) + \frac12 \sum_i \delta_i^2 \leq \frac12 (b^2 - a^2) + n (\max \delta_i)^2$$
So I need $\max \delta_i < \sqrt{\epsilon/n}$. If I take an equally spaced partition, then $\max\delta_i = (b-a)/n$ and so I need $n > (b-a)^2/\epsilon$, which I can certainly do. This proves that $\inf S(P) = \frac12 (b^2 - a^2)$.