# Let $f(x)$ be a continuous positive function on $[1, 2]$. Area of the region $[1,2]$: (list of options below)

Let $f(x)$ be a continuous positive function on $[1, 2]$. Area of the region $[1,2]$: (list of options below) Here is the list I have to choose from:

$$a)\lim_{n\to\infty} \sum_{i=1}^{n} f(1 + \frac{i-1}{n})\frac{1}{n}$$ $$b)\lim_{n\to\infty} \sum_{k=0}^{n-1} f(1 + \frac{k-1}{n})\frac{1}{n}$$ $$c)\lim_{n\to\infty} \sum_{i=1}^{n} \frac{f(1 + \frac{i}{n})}{n}$$ $$d) \int_{1}^{2} f(x) dx$$ $$e) \text { All of the above}$$ $$f) \text { None of the above}$$

From the list above I know that $\Delta x = \frac{1}{n}$, and so $x_i^* = a + i\Delta x = 1 + \frac{i}{n}$. I know that $d$ is true. I also know that $c$ is True because it follows the format $b)\lim_{n\to\infty} \sum_{i=0}^{n} f(1 + i\Delta x)\Delta x$.

Since $c$ and $d$ are true, then all must be true, but can someone explain why $a$ and $b$ are true?

## 1 Answer

For example, $$\sum_{i=1}^{n} f\left(1 + \frac{i-1}{n}\right)\frac{1}{n}=\sum_{i=1}^{n} \frac{f(1 + \frac{i}{n})}{n}+\frac{f(1)-f(2)}n.$$ Since $$\lim_{n\to\infty} \frac{f(1)-f(2)}n=0$$ the limits in a) and c) are the same. You can use a similar argument to show that these also coincide with the limit in b).