0

Hi Guys I am trying to understand what is going on as it relates to the range of the X values since from the question, the range is stated in terms of a and b. I have been exposed to a few question where the closed interval is give example([0,1] or [0,4]) but for this question I am not sure how to evaluate this function and determine if it is Riemann Integrable. Can anyone help me.

enter image description here

John
  • 369
  • $\int_{0}^2f(x)dx=2b$ We shouldn't concern ourselves with this one point (when $x=1$) in the interval of integration where $f(x)=a$ because that doesn't really change the area under $f(x)$. This should match your intuition of what an integral means. Now can we demonstrate that $f$ is (Riemann) integrable? What criterion was given in the text that your exploring on the topic? – Mason Oct 02 '19 at 20:22
  • The criterion which we are looking at is to see if the function is Riemann integrable is the comparison of the Upper and Lower Darboux sum @Mason – John Oct 02 '19 at 20:26
  • Beautiful. So you have a function which is constant $b$ and then has this one annoying point $(1,a)$ does this mean that your lower and upper darboux sums will converge to different limits as we let the partitions become finer and finer? – Mason Oct 02 '19 at 20:33
  • no @Mason, what is confusing me is using the Lower and Upper Darboux sum to show if its indeed riemann integrable based on the unknown range. I do understand what you are saying. Essentially evaluation of both the Lower and Upper sums is where the confusion is. – John Oct 02 '19 at 20:36
  • I am not sure I understand your meaning with "unknown" range. The range is given: The function evaluates to either $a$ or $b$. You mean we don't know what $a$ and $b$ are? We should just interpret them as some constant right? – Mason Oct 02 '19 at 20:47
  • yes @Mason just not to sure how to evaluate the upper and lower sum integrals when it is defined as arbitrary constants. – John Oct 02 '19 at 21:10

1 Answers1

0

Let's review what these Darboux Sums mean.

$$\begin{align} L(f, P) &= \sum_{i=0}^{n-1} \inf_{t \in [x_i, x_{i+1}]} f(t)(x_{i+1} - x_i), \\ U(f, P) &= \sum_{i=0}^{n-1} \sup_{t \in [x_i, x_{i+1}]} f(t)(x_{i+1} - x_i). \end{align}$$ Ok. So let's pick a partition $P$ of $[0,9]$ And compute these sums.

Let $t_0=0, t_1=1/7, t_2=2/7 \dots t_{k}=k/7, t_{63}=9$. Then $P=[t_0 \dots t_{63}]$. When we are computing these Darboux Sums we look at the interval

$[5/7, 6/7]$ and we think $f(x)$ on this interval has what infimum value? The answer is $b$. Along $[5/7, 6/7]$ this function has only one value: $b$.

So we should evaluate $$\inf_{t\in [t_5,t_6]}f(t) \times 1/7=b/7$$ To make progress we have to make an assumption about $b-a$ is this value positive or negative or zero? Let's assume $a<b$. Then $L(f,P)=\frac{61}{63}b+\frac{2}{63}a$. There are these two intervals where the $\inf$ of our function evaluates to $a$. Namely and [$[t_6,t_7]=[\frac{6}{7}, 1]$ and $[t_7,t_8]=[1,\frac{8}{7}]$. Note that the $U(f,P)=b$.

So this is how to evaluate this thing on a specific partition.

Mason
  • 3,792