Evaluate $\int_0^3 x{\sqrt {3-x}}\;dx $ using Riemann sums

Question

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

Answer 1

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 $$\lim_{n\to\infty}\frac{9\sqrt3}{n^{5/2}}s_n=\frac{12\sqrt3}5.$$