2
$\begingroup$

I want to prove the first FTC:

$$F(b) = \int_0^bf(x)dx$$

Using the following deffinition of integral:

$$\int_a^bf(x)dx = \lim_{n \to \infty } \sum_{i=0}^nf( \Delta x i) \Delta x$$

where $$ \Delta x = \frac {b-a}{n}$$

I have been trying to get it right but I'm missing something, how can I go around to prove it using this defintion of integral?

$\endgroup$
  • 1
    $\begingroup$ you can't have x both as one of the limits and as the variable to integrate with respect to. you can switch for example to $f(t)dt$ if you wish $\endgroup$ – mathreadler Mar 6 '17 at 19:31
  • $\begingroup$ i think i corrected that a few seconds ago, just replaced x with b. is that what you meant? $\endgroup$ – Joaquin Brandan Mar 6 '17 at 19:33
  • $\begingroup$ Maybe you fixed the second but you seem to have forgot to fix it in the first equation $\endgroup$ – mathreadler Mar 6 '17 at 19:33
  • $\begingroup$ $F(b)$ then. The thing about the theorem is that the variable of the function is in the upper limit. Yes that also works. $\endgroup$ – mathreadler Mar 6 '17 at 19:36
  • $\begingroup$ Sorry for the mess, i'm trying to get variable names that match both equations so that i can wrap my head around what's going on. does it make sense now? $\endgroup$ – Joaquin Brandan Mar 6 '17 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.