In my homework we start out with
$$\int_{x=0}^3 x^2 \, dx=\lim_{P: \Delta x \to 0} \sum_{i = 1}^n f(x_i) (\Delta x)_i$$
Where I take
$$P_i=[\frac{i-1}{n},\frac{i}{n}], x_i=\frac{i}{n}, (\Delta x)_i=\frac{3}{n}$$
So then
$$\int_{x=0}^3 x^2 \, dx=\lim_{n \to \infty} \sum_{i = 1}^n \frac{i^2}{n^2} \frac{3}{n}$$
With given in my homework
$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$
Makes
$$\int_{x=0}^3 x^2 \, dx=\lim_{n \to \infty} \frac{n(n+1)(2n+1)}{2n^3}=\lim_{n \to \infty} \frac{2n^3+3n^2+n}{2n^3}$$
Which leads me to
$$\int_{x=0}^3 x^2 \, dx=\lim_{n \to \infty} 1+\frac{3}{2n}+\frac{1}{2n^2}$$
This is not the answer I should be getting… Should I choose $P_i, x_i, (\Delta x)_i$ differently? Am I going wrong somewhere else?
Best Answer
Yes, you have a mismatch in the $P_i, x_i, (\Delta x)_i$.
For, we always have that for $P_i = [r_1,r_2]$, $(\Delta x)_i = r_2 -r_1$. Another problem is that:
$$\bigcup_{i=1}^n P_i = [0,1] \ne [0,3]$$
Can you see how to adjust the $P_i$ and $x_i$ to solve both these issues in one go?