Evaluating $\int_{0}^{3} \sqrt{1+x}\: dx$ using Limit of a Sum approach Evaluate $\int_{0}^{3} \sqrt{1+x}\: dx$ using Limit of a Sum approach.
Using the formula $$\int_{a}^{b} f(x)\:dx=(b-a) \times \lim_{n \to \infty} \frac{1}{n} \times \sum_{k=1}^{n}f\left(a+\frac{(b-a)k}{n}\right)$$ we have
$$I=\int_{0}^{3} f(x)\:dx=(3-0) \times \lim_{n \to \infty} \frac{1}{n} \times \sum_{k=1}^{n}f\left(0+\frac{(3-0)k}{n}\right)=3 \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f\left(\frac{3k}{n}\right)$$ So
$$I=3\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n} \sqrt{1+\frac{3k}{n}}$$ Now if we expand the summation we get limits of the form
$$\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{3}{n}}+\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{6}{n}}+\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{9}{n}}+\cdots+\lim_{n \to \infty} \frac{1}{n}\sqrt{1+\frac{3n}{n}}$$
But each limit is clearly zero and hence $I=0$.  I know the answer is wrong but what is my mistake?
 A: Consider 
$$\int_0^3 \sqrt{1+x} \, dx = \lim_{n \to \infty} \frac{3}{n}\sum_{k=1}^n\sqrt{1 + \frac{3k}{n}} = \lim_{n \to \infty} \frac{3}{n^{3/2}}\sum_{k=1}^n\sqrt{n + 3k}.$$
Using the binomial expansion, we have 
$$(n + 3k - 3)^{3/2} = (n + 3k)^{3/2} \left(1 - \frac{3}{n+3k} \right)^{3/2} \\ = (n + 3k)^{3/2}\left(1 - \frac{3}{2}\frac{3}{n+3k}  + O(1/n^2)\right),$$
and
$$\sqrt{n+3k} = \frac{2}{9}\left[(n+3k)^{3/2} - (n+3k-3)^{3/2}\right] +O(1/\sqrt{n}).$$
Summing we get
$$\lim_{n \to \infty}\frac{3}{n^{3/2}}\sum_{k=1}^n\sqrt{n + 3k} = \lim_{n \to \infty}\frac{3}{n^{3/2}}\sum_{k=1}^n\frac{2}{9}\left[(n+3k)^{3/2} - (n+3k-3)^{3/2}\right] \\ = \lim_{n \to \infty}\frac{3}{n^{3/2}}\frac{2}{9}\left[(4n)^{3/2} - (n)^{3/2}\right] \\ = \frac{2}{3}\left[4^{3/2} - 1\right].$$
A: You have to choose the partition points $x_k$ such that $\sqrt{1+x_k}$ becomes manageable. Therefore put
$$x_k:=u_k^2-1\quad(0\leq k\leq n)$$ whereby the $$u_k:=1+{k\over n}\quad(0\leq k\leq n)$$
are equally spaced. The $x_k$ are then inequally spaced between $0$ and $3$, but in any case the differences $x_k-x_{k-1}$ tend to $0$ when $n\to\infty$. A typical Riemann sum is then given by
$$\eqalign{R_n&:=\sum_{k=1}^n \sqrt{1+x_k}\>(x_k-x_{k-1})=\sum_{k=1}^n u_k(u_k+u_{k-1})(u_k-u_{k-1})\cr &={1\over n}\sum_{k=1}^n\left(1+{k\over n}\right)\left(2+{2k-1\over n}\right)={14\over3}+{3\over2n}-{1\over 6n^2}\ .\cr}$$
Note that we have computed the last sum exactly, using the formulas for $\sum_{k=1}^n k^\alpha$ when $\alpha\in\{0,1,2\}$. It follows that
$$\int_0^3\sqrt{1+x}\>dx=\lim_{n\to\infty} R_n={14\over3}\ .$$
