Transforming a Riemann Sum to an integral.

Question

Question. Consider the limit $$\begin{align} L&=\lim\limits_{n\to\infty}\sum_{k=1}^n\dfrac{k\sqrt{n+k}}{n^{5/2}}\\&=\lim\limits_{n\to\infty}\left(\dfrac{\sqrt{n+1}}{n^{5/2}}+\dfrac{2\sqrt{n+1}}{n^{5/2}}+\dfrac{3\sqrt{n+1}}{n^{5/2}}+\cdots+\dfrac{n\sqrt{n+1}}{n^{5/2}}\right) \end{align}$$

(a) $L$ is a definite integral, that is $L=\int_a^bf(x)\,\mathrm dx$, for some function $f$, and some numbers $a$ and $b$. Find $f(x)$, $a$ and $b$.

I could not transform Riemann sum to integral.

Answer 1

Note: $$\sum\frac{k\cdot \sqrt{n+k}}{n^{5/2}}=\sum\frac{1}{n} \cdot\frac{k}{n} \cdot\sqrt{1+\frac{k}{n}}$$

Now this is of the form $\sum\frac{1}{n}\cdot f(\frac{k}{n}) = \int_{0}^{1}f(x) \ dx$

Answer 2

According to the definition of definite integral if $y=f(x)$ be a continuous function on interval $[a,b]$ then $$\int^a_bf(x)dx=\lim_{\Delta x\rightarrow0}\sum_{x=a}^bf(x)\Delta x$$ In a special numerical methods, based on dividing the interval into $n$ equal parts of lenght, we get $\Delta x=(b-a)/n$. So $$\int^a_bf(x)dx=\lim_{n\rightarrow\infty}\underbrace{\frac{b-a}{n}}_{B}\sum_{k=1}^n\underbrace{f\bigg(a+\frac{k(b-a)}{n}\bigg)}_{A}$$ Now, we have $$\frac{k\sqrt{n+k}}{n^{5/2}}=\underbrace{\frac{1-0}{n}}_{B}\times\underbrace{\frac{k\sqrt{1+k/n}}{n}}_{A}=\frac{1}{n}f\bigg(0+\frac{k}{n}\bigg)$$ where $f(x)=x\sqrt{1+x}$.