What integral does the Riemann sum $\frac{1}{30}\sum_{k=1}^{60}e^{k/30}$ approximate? 
What integral does the Riemann sum $\frac{1}{30}\sum_{k=1}^{60}e^{k/30}$ approximate?


Can someone please tell me what steps I need to follow to solve this problem? I know that the answer is B but I don't understand how the interval changed to $[0,2]$ and the change in x of $(1/30)$ suddenly disappeared in the process.
 A: Hint:
$$\lim_{n \to \infty} \frac1{n} \sum_{k=1}^{\color {red} n} f(k/n) = \int_0^1 f(x) dx$$
A: A Riemann sum for
$$\int_a^b f(x)\,dx$$
using $n$ subintervals is
$$\frac{b-a}n\sum_{k=1}^n f(x_k)\ ,$$
where $x_k$ is a point in the $k$th subinterval.  Compare this with the given sum
$$\frac1{30}\sum_{k=1}^{60} e^{k/30}\ .$$
By looking at the upper limit of summation, $n=60$.  Then from the fraction at the front, $b-a=2$, and since all your answer options have $a=0$, it follows that $b=2$.  Finding the function is a little harder as the exact value of $x_k$ is not specified - it could be the least $x$ value in the $k$th subinterval, or the greatest, or anything in between.  However, as the $k$th subinterval is
$$a+(k-1)\frac{b-a}n\le x\le a+k\frac{b-a}n\ ,$$
that is in this case
$$\frac{k-1}{30}\le x\le\frac{k}{30}\ ,$$
we have $x\approx\frac{k}{30}$, so we are looking, roughly, at
$$f\Bigl(\frac{k}{30}\Bigr)=e^{k/30}\ ,$$
which looks like $f(x)=e^x$.  So it seems that the integral should be
$$\int_0^2 e^x\,dx\ ,$$
and you can check by finding a Riemann sum for this integral, using $60$ subintervals, confirming that you get the given sum.
