Proving x^2 + x is integrable.

[STEMucator]

Homework Statement

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{s_p} = inf{S_p} = J.

Homework Equations

s_p is my underestimate and S_p is my overestimate.

The Attempt at a Solution

So first off we note :

m_i = inf{ f(x) | x in [2,5] } = 6
M_i = 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?

Thanks in advance.

 

[Ray Vickson]

Homework Statement

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{s_p} = inf{S_p} = J.

Homework Equations

s_p is my underestimate and S_p is my overestimate.

The Attempt at a Solution

So first off we note :

m_i = inf{ f(x) | x in [2,5] } = 6
M_i = 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?

Thanks in advance.

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) ,\;\;\;\;\; <br /> M_i = \sup_{x_i \leq x \leq x_i + \Delta x_i} f(x), \; i = 1,2 \ldots, n .

 

[pasmith]

Homework Statement

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{s_p} = inf{S_p} = J.

Homework Equations

s_p is my underestimate and S_p is my overestimate.

The Attempt at a Solution

So first off we note :

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

This is your error.

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

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

Since these limits are independent of \max \delta_i, we must have
<br /> \sup s(P) \leq \frac13(5^3 - 2^3) + \frac12 (5^2 - 2^2) \leq \inf S(P)<br />
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) <br /> + \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.

 

[STEMucator]

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{ S_p } ≤ S_p and since inf{ S_p } 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 sort of what I wanted to happen, except I want 99/2, not 25/2.

Did i partition incorrectly?

 

[pasmith]

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

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 &gt; 0 there exists a paritition P such that
\frac 12(b^2 - a^2) &lt; S(P) &lt; \frac 12(b^2 - a^2) + \epsilon
So I have
S(P) = \sum_i x_i (x_i - x_{i-1}) <br /> = \sum_i \frac{x_i + x_{i-1} + x_i - x_{i-1}}{2}(x_i - x_{i-1})<br /> = \sum_i \frac12 (x_i^2 - x_{i-1}^2) + \sum_i \frac12 (x_i - x_{i-1})^2 \\<br /> = \frac12 (b^2 - a^2) + \frac12 \sum_i \delta_i^2 <br /> \leq \frac12 (b^2 - a^2) + n (\max \delta_i)^2<br />
So I need \max \delta_i &lt; \sqrt{\epsilon/n}. If I take an equally spaced partition, then \max\delta_i = (b-a)/n and so I need n &gt; (b-a)^2/\epsilon, which I can certainly do. This proves that \inf S(P) = \frac12 (b^2 - a^2).