I was wondering what happens to the constant $\pi$ when we change from the riemansum,
$\lim_{n \to \infty}\sum\limits_{i=1}^n \dfrac{\pi}{n} \sin(\dfrac{{\pi}i}{n})$
to an integral form. My guess is that we end up with the integral
$\int_{0}^{\pi} \sin({\pi}x)$ and not $\int_{0}^{\pi} \sin(x)dx$
My books disagrees while symbolab agrees so I'm rather confused. Anyone have any enlightenment?
You should probably use the following formula :
$$ \lim _{n \rightarrow+\infty} \frac{b-a}{n} \sum_{k=1}^n f\left(a+k \frac{b-a}{n}\right)=\int_a^b f(t) d t $$
By choosing $a=0, b=\pi$, we get
$$ \lim _{n \rightarrow+\infty} \frac{\pi}{n} \sum_{k=1}^n f\left(k \frac{\pi}{n}\right)=\int_0^\pi f(t) d t $$
Then we choose $f(x)=\sin(x)$ :
$$ \lim _{n \rightarrow+\infty} \frac{\pi}{n} \sum_{k=1}^n \sin\left(k \frac{\pi}{n}\right)=\int_0^\pi \sin(t) d t=2 $$
Note that $\int_0^\pi \sin(t) d t =\pi\int_0^1 \sin(\pi t) d t$, maybe that's what you found.