Question

Need help with an integration question below! :) Thanks!

Answer 1

The limit definition of a definite integral, right?

Answer 2

@ParthKohli I believe so, yes

Answer 3

That'd be\[\lim_{n \to \infty}\sum_{r=1}^{n} \frac{1}nf\left(\frac{r}n\right)\]

Answer 4

Can you use that?

Answer 5

Just to be clear, is that \(e\) an arbitrary constant or *the* constant?

Answer 6

Latter

Answer 7

@ParthKohli I've asked that question, but apparently I can't use that :(

Answer 8

Would it be possible to prove it through Riemann sums?

Answer 9

What is the precise definition of the definite integral?

Answer 10

\[\int\limits_{b}^{a}f(x)dx=\lim_{n \rightarrow \infty}\sum_{i=1}^{n}f(x_1^*)(\Delta x)\]

Answer 11

To me, the set [0,1] has volume, and f has only one discontinuity , at x = 1/3. Since f is bounded, f is integrable.
:) Real Analysis is never easy. Good luck.

Answer 12

* x= e/3

Answer 13

ITs like 4 0r 2

Answer 14

Is this valid?

There are two cases. In the Riemann sum:
1) there exists one and only one term for which \(f(x_1^*)= \frac e 3\). In this case:
\[ \lim_{n \rightarrow \infty}\sum_{i=1}^n f(x_1^*)\Delta x = \lim_{n \rightarrow \infty} \frac e 3 \Delta x = 0\]
We can justify this move by using some theorem about shifting sequences and by using the proper limits definitions if you want to be very rigorous.
2) Else, \(f(x_1^*)=0\) for all terms and so trivially the Riemann Sum is 0.

Answer 15

@Bobo-i-bo makes sense! Thanks so much! :)

Answer 16

@Bobo-i-bo but what about when x=e/3 and is equal to 1? It wouldn't be 0 then, would it?

Answer 17

Oops, i made an error in the first case. It should say "one and only one term for which \(f(x_i^*)=1\)" (in other words, when \(x_i^*=\frac e 3\)) and not \(f(x_i^*)=\frac e 3\)