Riemann sum of $\sin x$ on [0,1]

Question

For each positive integer $n$, define a function $f_n$ on [0,1] as follows:

$f_n(x)$= $0$ $\forall$ $x=0$

$\sin$($\pi\over{2n}$) $\forall x\in(0,{1\over{n}}]$

$\sin$($2\pi\over{2n}$) $\forall x\in({1\over{n}},{2\over{n}}]$

$\sin$($3\pi\over{2n}$) $\forall x\in({2\over{n}},{3\over{n}}]$

. . .

. . .

. . .

$\sin$($n\pi\over{2n}$) $\forall x\in({n-1\over{n}},1]$

Then, find $\lim_{n\to\infty}$ $\int_0^1f_n(x)dx$

My approach: I think it has to be converted into riemann integral. However I am not being able to do that.

Answer 1

This should get you started.

$\int_0^1f_n(x)dx$ is the right reimmann sum for the function $\sin{(\frac{\pi}{2} x)}.$ This follows since $$\int_0^1f_ndx = \sum_{i=0}^{n-1} \int_{\frac{i}{n}}^{\frac{i+1}{n}}f_ndx = \sum_{i=0}^{n-1} \frac{1}{n} sin(\frac{(i+1) \pi}{2n}).$$ Hence, the limit as $n \longrightarrow \infty$ of the expressions above is actually equal to the integral:

$$\int_0^1 \sin{\frac{\pi}{2}x}dx$$

I have left some details for you to work out.

but cheers!