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Let $f:[0,2] \rightarrow \mathbb{R}$ be given by

$$f(x) = \left\{ \begin{array}\ 10,\quad \ 0\leq x < 1, \\ 100,\quad \ x = 1, \\ -5,\quad \ 1 < x \leq 2. \\ \end{array} \right. $$

Prove that $f$ is Darboux integrable and compute $\int_{0}^{2}f$.

Attempt

That the function is Darboux integrable means for all $\epsilon > 0$, there exists a partition $P$ of $[0,2]$ such that $U(f,P) - L(f,P) < \epsilon$.

Suppose $P = \{t_{0}, \dots, t_{n}\}$ is a partition of $[0,2]$ with $t_{j-1} < 1 < t_{j}$.

For, $t_{0} < \dots < t_{j-1}$: $m_{i} = M_{i} = 10$

Also for $t_{j} < \dots < t_{n}$: $m_{i} = M_{i} = -5$

Now I'm having trouble with managing $x = 1$ which is where the discontinuity is and obviously the challenge of the problem. At first I was going to say that $m_{i} = M_{i} = 100$ whichever interval the number 1 is in and that would give me:

$$L(f,P) = \sum_{i = 1}^{n}m_{i}(t_{i} - t_{i - 1}) = \sum_{i = 1}^{j-1}10(t_{i} - t_{i-1}) + 100(t_{j} - t_{j-1}) + \sum_{j+1}^{n}-5(t_{i} - t_{i-1})$$

and

$$U(f,P) = \sum_{i = 1}^{n}M_{i}(t_{i} - t_{i - 1}) = \sum_{i = 1}^{j-1}10(t_{i} - t_{i-1}) + 100(t_{j} - t_{j-1}) + \sum_{j+1}^{n}-5(t_{i} - t_{i-1})$$

Then subtracting these I would get $U(f,P) - L(f,P) = 0 < \epsilon$.

But I feel this isn't the right answer and I need to express the partition a bit more explicitly. Also whatever $U(f,P) - L(f,P)$ is, it is going to eventually converge to what the integral is. And when calculating the integral (as a check using earlier calc techinques) I get $5$ which is not the value of the difference in the upper and lower sums. Where am I going wrong?

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    $\begingroup$ just a comment. if $[t_{j-1}, t_j]$ is the interval that contains one, with $t_j-1 < 1 < t_j$ , then $M_j = 100$ and $m_j = -5$, thats because the interval will have elements belonging to $[1,2]$ $\endgroup$ Aug 16 '20 at 4:56
  • $\begingroup$ Also. see this math.stackexchange.com/questions/857961/… $\endgroup$ Aug 16 '20 at 5:03
  • $\begingroup$ U and L converge to what the integral is, not U-L, which converges to 0? $\endgroup$ Aug 16 '20 at 5:13
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Fixed $\varepsilon$ and taken $\delta$ small enough, you can take $$ t_{j-1}=1-\delta,\qquad t_j=1+\delta, $$
so that \begin{align} L(f,P) &= \sum_{i = 1}^{n}m_{i}(t_{i} - t_{i - 1}) = \sum_{i = 1}^{j-1}10(t_{i} - t_{i-1}) + \mathbf{(-5)}(t_{j} - t_{j-1}) + \sum_{j+1}^{n}(-5)(t_{i} - t_{i-1})=\\ &=10(1-\delta)-5\cdot2\delta-5(1-\delta)=5-15\delta\\ U(f,P) &= \sum_{i = 1}^{n}M_{i}(t_{i} - t_{i - 1}) = \sum_{i = 1}^{j-1}10(t_{i} - t_{i-1}) + 100(t_{j} - t_{j-1}) + \sum_{j+1}^{n}(-5)(t_{i} - t_{i-1})=\\ &=10(1-\delta)+100\cdot2\delta-5(1-\delta)=5+195\delta \end{align} so that to have $$ U(f,P)-L(f,P)=210\delta<\varepsilon $$ you have to choose $$ \delta<\frac{\varepsilon}{180}. $$

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  • $\begingroup$ Hi. I was trying to figure out how you got the line $10(1-\delta)-5\cdot2\delta-5(1-\delta)=5-15\delta$. I was working it out and I see that it is supposed to be a telescoping sum of sorts, but I end up with the following if I'm doing it right: $10(1+\delta - (1-\delta)) + (-5)((1+\delta - (1-\delta)) + (-5)(1+\delta - (1-\delta))$ $\endgroup$
    – dc3rd
    Aug 16 '20 at 17:46
  • $\begingroup$ No telescoping, The first sum gives $m_i=10$ times the complessive amplitude of the left intervals, i.e. $1-\delta$, the second the minimum, i.e. $-5$ times the amplitude of the central interval, i.e. $2\delta$, the third $m_i=-5$ times the complessive amplitude of the right intervals, i.e. $1-\delta$ $\endgroup$
    – enzotib
    Aug 16 '20 at 17:52
  • $\begingroup$ I see, but wouldn't the interval be $t_{i} - t_{i-1} = ((1+\delta) - (1- \delta))$ ? That seems to be where my misunderstanding is $\endgroup$
    – dc3rd
    Aug 16 '20 at 17:54
  • $\begingroup$ @dc3rd yes, $t_j-t_{j-1}=(1+\delta)-(1-\delta)=2\delta$, for the central interval, but $$\sum_{i=1}^{j-1}(t_i-t_{i-1})=t_{j-1}-t_1=(1-\delta)-0=1-\delta.$$ Similarly on the righ you have $2-(1+\delta)=1-\delta$ $\endgroup$
    – enzotib
    Aug 16 '20 at 18:01
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    $\begingroup$ @dc3rd Yes you're right on both, corrected $\endgroup$
    – enzotib
    Aug 16 '20 at 18:56

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