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?

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
$$\lim_{n \to \infty}\frac{3}{n^{3/2}}\sum_{k=1}^n\sqrt{n + 3k} = \lim_{n \to \infty}\frac{3}{n^{3/2}}\sum_{k=1}^n\frac{2}{9}\left[(n+3k)^{3/2} - (n+3k-3)^{3/2}\right] \\ = \lim_{n \to \infty}\frac{3}{n^{3/2}}\frac{2}{9}\left[(4n)^{3/2} - (n)^{3/2}\right] \\ = \frac{2}{3}\left[4^{3/2} - 1\right].$$
