What is the limit of given the series? Find the limit of this series.
$\lim_{n \to \infty} \sum_{i=1}^n {{i \sqrt{n^2 - i^2}}  \over n^3}$

Here is my attempt)
$\lim_{n \to \infty} \sum_{i=1}^n ({i \over n})\sqrt {{1- {({i \over n})^2}}}$ 
= $\int_0^1 xi \sqrt{1-(ix)^2}$
= $\int_0 ^1 xi \sqrt{1+x^2}$
= $i$ [${1 \over 3}(1+x^2)^{3\over2} $]$_{0} ^ 1$
= $i \over 3$ $(2 ^ {3 \over 2} -1)$
But the answer was $1 \over 3$. I don't know which point I have a mistake.
Any help would be appreciated.
 A: As $\sqrt{\dfrac1{n^2}}=\dfrac1n$ for $n>0,$
$$\dfrac{i\sqrt{n^2-i^2}}{n^3}=\dfrac1n\cdot\dfrac in\sqrt{1-\left(\dfrac in\right)^2}$$
Now use The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$  and replace $\sqrt{1-x^2}=y\implies1-x^2=y^2$
A: It's $$\lim_{n\rightarrow+\infty}\sum_{i=1}^n\left(\frac{i}{n}\sqrt{1-\frac{i^2}{n^2}}\frac{1}{n}\right)=\int_0^1x\sqrt{1-x^2}dx=\frac{(1-x^2)^{\frac{3}{2}}}{\frac{3}{2}\cdot(-2)}|_0^1=\frac{1}{3}.$$
A: If $b=1,a=0$
$$\int_{0}^{1}f(x)dx=\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=1}^{n}f\left(\dfrac{i}{n}\right)\tag{1}$$
So to convert this limit $\lim_{n\to\infty}\sum_{i=1}^{n}\dfrac{i\sqrt{n^2-i^2}}{n^3}$ in the form of equation $(1)$, we rewrite it as follows:-
$$\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=1}^{n}\dfrac{i}{n}\sqrt{1-\dfrac{i^2}{n^2}}$$
So we can say $f\left(\dfrac{i}{n}\right)=\dfrac{i}{n}\sqrt{1-\dfrac{i^2}{n^2}}$
Replacing $\dfrac{i}{n}$ with $x$
$$f(x)=x\sqrt{1-x^2}$$
$$\int_{0}^{1}f(x)dx=\int_{0}^{1}x\sqrt{1-x^2}dx$$
$$1-x^2=t$$
$$-2x=\dfrac{dt}{dx}$$
$$xdx=-\dfrac{dt}{2}$$
$$\int_{0}^{1}f(x)dx=\dfrac{1}{2}\cdot\int_{0}^{1}t^\frac{1}{2}dt$$
$$\int_{0}^{1}f(x)dx=\dfrac{1}{3}$$
Hope it will help you in understanding why we integrals are helpful in such cases.
