One of my homework questions goes:
The following sum $$\sqrt{5 + \frac{4}{n}} \cdot \left( \frac{4}{n}\right) + \sqrt{5 + \frac{8}{n}} \cdot \left( \frac{4}{n}\right) + \ldots + \sqrt{5 + \frac{4 n}{n}} \cdot \left( \frac{4}{n}\right)$$ is a right Riemann sum for the definite integral $$\int_{4}^b f(x)\, dx$$ It is also a Riemann sum for the definite integral $$\int_{5}^c g(x)\, dx$$ I need to find $b, f(x), c$ and $g(x)$. So far I've figured out that b=8 and c=9, but am not sure about $f(x)$ or $g(x)$. If $\Delta x$ is $1+ \frac4n$ for the first integral, I figure $f(x)=\sqrt{4+x}$ and following that logic $g(x)=\sqrt{x}$. Am I headed in the right direction? (I only have one attempt left so I need to be sure, any help is hugely appreciated)
The $\Delta x$ won't be $1 + \frac{4}{n}$. The $\Delta x$ should tend to $0$ as $n$ tends to $\infty$, but $1 + \frac{4}{n}$ is always greater than $1$.
If we partition $[4, 8]$ into $n$ uniform intervals, then the length of each interval, which is $\Delta x$, will be $\frac{8 - 4}{n} = \frac{4}{n}$. The $n + 1$ endpoints of these intervals will be: $$4, 4 + \Delta x, 4 + 2\Delta x, 4 + 3\Delta x, \ldots, 4 + n\Delta x = 8,$$ i.e. $$4, 4 + \frac{4}{n}, 4 + \frac{2 \cdot 4}{n}, 4 + \frac{3 \cdot 4}{n}, \ldots, 4 + \frac{n \cdot 4}{n} = 8.$$ If we are taking the right Riemann sums, we just ignore the first endpoint $4$. So, whatever our function $f$ is going to be, our Riemann sum will take the form, $$f\left(4 + \frac{4}{n}\right) \frac{4}{n} + f\left(4 + \frac{2 \cdot 4}{n}\right) \frac{4}{n} + \ldots + f\left(4 + \frac{n \cdot 4}{n}\right) \frac{4}{n}.$$ Our actual Riemann sum looks like this: $$\sqrt{5 + \frac{4}{n}} \frac{4}{n} + \sqrt{5 + \frac{2 \cdot 4}{n}} \frac{4}{n} + \ldots + \sqrt{5 + \frac{n \cdot 4}{n}} \frac{4}{n}.$$ The choice $f(x) = \sqrt{1 + x}$ seems natural here.
See if you can adapt this for the interval $[5, 9]$ instead.