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

Question

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.

Answer 1

Hint:

$$\lim_{n \to \infty} \frac1{n} \sum_{k=1}^{\color {red} n} f(k/n) = \int_0^1 f(x) dx$$

Answer 2

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.