So, I was asked to do this integral using the limit method (or the Riemann Sum)
$$\int_0^3 x{\sqrt {3-x}}\;dx $$
And, I do it like this:
$$\int_0^3 x{\sqrt {3-x}}\;dx $$
Firstly, I determine the $\Delta x$ and $c_i$
$$\Delta x = \frac{b-a}{n}$$
$$\Delta x = \frac{3-0}{n}$$
$$\Delta x = \frac{3}{n}$$
Using the right end point :
$$c_i=a+\Delta xi$$
$$c_i=0+\frac{3}{n}i$$
$$c_i=\frac{3i}{n}$$
Then, evaluating the Integral using limit:
$$\lim_{n\to\infty}\;\sum_{i=1}^n\;f(c_i)\Delta x $$
$$\lim_{n\to\infty}\;\sum_{i=1}^n\;f\left(\frac{3i}{n}\right)\left(\frac{3}{n}\right)$$
$$\lim_{n\to\infty}\;\sum_{i=1}^n\;\left(\frac{3i}{n}{\sqrt {3-\frac{3i}{n}}}\right)\;\left(\frac{3}{n}\right)$$
$$\lim_{n\to\infty}\;\left(\frac{9 \sqrt {3}}{n^\frac {5}{2}}\right)\;\sum_{i=1}^n\;i{\sqrt {n-i}}$$
And, now I'm stuck here. Is there a way to do a summation that has a square root in it?
notes : if you find error in my calculation please let me know
Here are three more or less equivalent methods to get rid of the square root and solve your problem. They all rely on the fact that for $p=2$ or $4$, $$\lim_{n\to\infty}\frac{\sum_{k=1}^nk^p}{n^{p+1}}=\frac1{p+1}.$$(Actually, this holds for every integer $p\ge0$, but can also be proved for each $p$ by direct computation of the numerator: $\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$ and $\sum_{k=1}^nk^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.)
First method: before taking Riemann sums as you were ordered to, first do a substitution $y=\sqrt{3-x}$ in your integral, i.e. apply the formula $$\int_{g(a)}^{g(b)}f(x)\,dx=\int_a^bf(g(y))g'(y)\,dy$$ with $$f(x)=x\sqrt{3-x},\quad g(y)=3-y^2,\quad a=\sqrt3,\quad b=0$$ or less formally: in $\int_0^3x{\sqrt {3-x}}\,dx$, replace $\sqrt{3-x}$ by $y$ (which goes from $\sqrt{3-0}$ to $\sqrt{3-3}$ when $x$ goes from $0$ to $3$) hence $x$ by $3-y^2=:g(y)$ and $dx$ by $g'(y)dy=-2ydy$. This results in $$\begin{align}\int_0^3x{\sqrt {3-x}}\,dx&=\int_{\sqrt3}^0(3-y^2)y(-2y)\,dy\\&=2\int^{\sqrt3}_0(3-y^2)y^2\,dy\\ &=2\lim_{n\to\infty}\frac{\sqrt3}n\sum_{k=1}^n\left(3-\left(\frac{k\sqrt3}n\right)^2\right)\left(\frac{k\sqrt3}n\right)^2\\ &=18\sqrt3\left(\lim_{n\to\infty}\frac{\sum_{k=0}^nk^2}{n^3}-\lim_{n\to\infty}\frac{\sum_{k=0}^nk^4}{n^5}\right)\\&=18\sqrt3\left(\frac13-\frac15\right)\\&=\frac{12\sqrt3}5. \end{align}$$
Second method:instead of a Riemann sum with intervals of equal sizes, take directly $x_i$ such that $$\sqrt{3-x_i}=\sqrt3\left(1-\frac in\right)\quad(i=1,2,\dots,n).$$ (My first idea was to let $\sqrt{3-x_k}$ be equal to $\sqrt3\frac kn$. That way, when $k\in\{0,1,\dots,n\}$, the smallest $x_k$ would have been $x_n=0$ and the largest one, $x_0=3$. But I soon realized I had to reverse the ordering, to obtain $x_0=0$ and $x_n=3$. This explains the choice $\sqrt{3-x_{n-k}}:=\sqrt3\frac kn$, i.e. $\sqrt{3-x_i}:=\sqrt3\frac{n-i}n=\sqrt3\left(1-\frac in\right)$.)
You thus get $x_i-x_{i-1}<\frac6n\to0$ and $$S_n=\sum_{i=1}^n(x_i-x_{i-1})f(x_i)=9\sqrt3\sum_{i=1}^n\frac{2(n-i)+1}{n^2}\left(1-\frac{(n-i)^2}{n^2}\right)\frac{n-i}n$$ The $+1$ in the numerator above disappears in the limit, and there only remains the same limit as in the first method: $$\lim_{n\to\infty}S_n=9\sqrt3\lim_{n\to\infty}\sum_{k=0}^n\frac{2k}{n^2}\left(1-\frac{k^2}{n^2}\right)\frac kn=\frac{12\sqrt3}5.$$
Third method: cut your $$s_n:=\sum_{i=0}^ni\sqrt{n-i}$$ into slices easy to approximate. For $i=0$ to $n$, $k:=\lfloor\sqrt{n-i}\rfloor\in\{0,1,\dots,\lfloor\sqrt n\rfloor\}$, and for every non negative integer $k$, the number of integers $i\le n$ such that $\lfloor\sqrt{n-i}\rfloor=k$ is $(k+1)^2-k^2=2k+1$. Therefore, $$s_n>\sum_{k=0}^{\lfloor\sqrt n\rfloor-1}(2k+1)(n-(k+1)^2)k$$ $$>2\sum_{k=0}^{\lfloor\sqrt n\rfloor-1}k^2(n-(k+1)^2)$$ and $$s_n<\sum_{k=0}^{\lfloor\sqrt n\rfloor}(2k+1)(n-k^2)(k+1)$$ $$<2\sum_{k=0}^{\lfloor\sqrt n\rfloor}(k+1)^2(n-k^2)$$ Since both these lower and upper bounds are asymptotically equivalent to $2\left(n\frac{\sqrt n^3}3-\frac{\sqrt n^5}5\right)=\frac4{15}n^{5/2}$, we conclude that $$\lim_{n\to\infty}\frac{9\sqrt3}{n^{5/2}}s_n=\frac{12\sqrt3}5.$$