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So I've spent a long while on this question and got the correct answer, but not in the way by which I was meant to. I need help understanding where to go from where I ended:

So first I considered the standard definition:

$ \int_{a}^{b}f_{(t)} dt=\lim_{n\to\infty}\sum_{k=1}^{n}f_{(x_i)}\Delta x $

Seeing as I already had the integral on hand, I was able to form another expression and equate the two like so:

$ f_{(\frac{(4\pi k)}{13})} (\frac{4\pi}{13n})= (\frac{1}{n})(cos (\frac{4\pi k}{13}))(\frac{13}{4\pi}) $

After simplification and a bit or manipulation, I got an expression for $ f_{(t)}$:

$ f_{(t)}=(cos (t))(\frac{13}{4\pi}) $

I'm fairly confident with everything to this point (Using a calculator to evaluate this integral gets the correct answer). It is from here that I am troubled. From my understanding of the question, we are now meant to present a Riemann's Sum, which is simple now that we have the function and the initial definition:

$ \int_{0}^{\frac{4\pi}{13}}f_{(t)} dt=\lim_{n\to\infty}\sum_{k=1}^{n}(cos (\frac{4\pi k}{13n}))(\frac{13}{4\pi})(\frac{4\pi}{13n}) $

And voila! Just like that, I'm going in a circle and can't find a way out. Every other approach I've taken so far has led me somewhere I've already been, so I suspect I'm missing something here.


  • $\begingroup$ See math.stackexchange.com/questions/469885/… $\endgroup$ – lab bhattacharjee Jan 21 '18 at 2:48
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    $\begingroup$ They are asking you to compute a limit of a sum by identifying it as a Riemann sum for some integral and then doing the integral using anti derivatives and the FTC (i.e. the usual way we compute integrals in calculus class.) $\endgroup$ – spaceisdarkgreen Jan 21 '18 at 3:04
  • $\begingroup$ @spaceisdarkgreen Omg. I didn't get that the question was just looking for that. In which case, all I had to do is just evaluate the definite integral by hand, like you said, and that was it. Thank you! $\endgroup$ – Yazan Jan 21 '18 at 4:04

Noting the limits, the function should be $\frac{13}{4\pi}\cos(x)$ and the integral will be

$$\int_{0}^{4\pi/13} \frac{13}{4\pi}\cos(x) dx = \frac{13 \sin(\tfrac{4\pi}{13})}{4\pi} $$


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