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Suppose we have been asked to evaluate the integral of a bounded function, $f$ over the closed interval $[a,b]$

Then, further suppose that $f$ is integrable i.e $sup \ L(f,P) = inf \ U(f,P)$ where the $sup$ and $inf$ are taken over all possible partitions $P$.

Then, how do we actually evaluate the integral by using a sequence of partitions, and why does that method work? (Suppose it's too difficult to consider the sup or inf over all partitions)

John
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  • Your previous question does not appear to be too different from this one. I had indicated in comments that way to evaluate the integral for the specific example you asked. Did you face any problems while trying that out? – Paramanand Singh Jan 15 '18 at 05:28
  • I've not yet understood why the norm needs to go to 0. I understand that it does, not why it needs to. If n approaches infinity and the norm goes to, say, 3, why is it that the limit of $U(f,P_n)$ does not equal the integral?

    We have that $L(f,P_n) <= Integral <= U(f,P_n)$, so if $n-> \ infinity$ and both the left and right limit approach the same thing, then the integral should be equal to that limit as well, regardless of whether the norm is going to 0 or not.

     – John Jan 15 '18 at 05:42
  • Why do you want the number of intervals to tend to $\infty $? That's not apriori, but rather a consequence of the fact that norm of partition tends to $0$. – Paramanand Singh Jan 15 '18 at 05:50
  • It's time now that you look deeper into these ideas of norm and related stuff. See this answer https://math.stackexchange.com/a/2047959/72031 – Paramanand Singh Jan 15 '18 at 05:54
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    @Saad: Only for a sequence of partitions $P_n$ where the norm (length of longest subinterval) goes to $0$ will $U(f,P_n)$ converge to the integral. I think you are trying to overcomplicate things. Even the most elementary and intuitive treatment of a Riemann integral starts with approximating the area under a curve as the sum of rectangular areas and then explaining that the approximation can be made closer and closer to the area/integral as the width of each rectangle shrinks to $0$. – RRL Jan 15 '18 at 06:09
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    The linked answer shows why norm tending to $0$ works. To see that when norm does not tend to $0$ it does not work, you can construct simple examples and check that it does not work. More specifically you can find examples where norm does not tend to $0$ and $U(f, P_n), L(f, P_n) $ tend to different limits. – Paramanand Singh Jan 15 '18 at 06:09

1 Answers1

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The Riemann integral defined as $I = \sup_P L(f,P) = \inf_P U(f,P)$ is equivalent to the integral defined as $I = \lim_{\|P\| \to 0} S(f,P)$ where $S(f,P)$ is a sum of the form $\sum_{j=1}^n f(c_j)(x_j - x_{j-1})$. When $f$ is Riemann integrable, that limit is unique regardless of how the tags $c_j \in [x_{j-1},x_j]$ are chosen or how the partition points are distributed as $\|P\| = \max_{1 \leqslant j \leqslant n} (x_j - x_{j-1}) \to 0$. This is proved here.

To evaluate the limit, pick partitions and tags as convenient. For example, we can chose $c_j = x_j = a + j(b-a)/n$, in which case the conditions $(b-a)/n = \|P\| \to 0$ and $n \to \infty$ coincide. Whence,

$$I = \int_a^b f(x) \, dx = \lim_{n \to \infty} \frac{b-a}{n} \sum_{j=1}^n f(a + j(b-a)/n).$$

Generally, for all but the simplest functions, it may be difficult if not impossible to compute the definite integral (analytically) as a limit of Riemann sums. The fundamental theorem of calculus is often the more expedient approach if, of course, we can find a closed form for the anti-derivative.

Some non-trivial computations with Riemann sums are are given for $xe^x$ and for a more difficult example.

RRL
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