Simple Integral Calculus Problem - Clarification added to the Riemann Sum Yesterday, a friend of mine sent me this question. We got stuck due to the intervention of the symbol $\epsilon$. It makes the problem quite confusing... It's not the kind of calculus question that I'm used to. Here is the question:
Let $f: \Bbb R \to \Bbb R$ be the function defined by $f(x) = x^2 \forall x \in \Bbb R$
Consider one uniform partition $P: a=x_0 < x_1 < x_2 < x_3 < ... < x_n = b$ of the interval $I = [a,b] \subset [0,+\infty)$.
Evaluate, in each situation, the Riemann sum $S(n):=\sum_{j=1}^n f(\epsilon_j)\Delta x_j$ and then evaluate $lim_{n\to +\infty}S(n)$.
i) $f(\epsilon_j)=max(f(x)/x\in I_j)$
ii) $f(\epsilon_j)=min(f(x)/x\in I_j)$
iii) $\epsilon_j$= midpoint of the subinterval $I_j=[x_{j-1}, x_j]$
I suppose that the problem tells us the "definition" of $\epsilon$ in iii), but I don't know if it's 100% right, since iii) should be something that we are going to evaluate, not a definition!
I hope you guys can help us with this problem that seems to be easy but got us stuck!
 A: Hint. 
i) and ii) $f(x) = x^2$ is an increasing function, hence if $x_{j-1} < x < x_{j}$, then $x_{j-1}^2 < x^2 < x_{j}^2$. So what is the maximum (resp. minimum) of $f$ in the closed interval $I_j = [x_{j-1}, x_{j}]$? In other words, what is $f(\epsilon_j)$ for the $\epsilon_j \in I_j$ that maximizes (resp. minimizes) $f$ there.
iii) The midpoint of two points $x_{j-1}, x_j$ is $\frac{x_{j-1}+x_{j}}{2}$.
Pay attention to the word uniform partition, this means the length of the n intervals $[x_{j-1}, x_j]$ is the same for all $j$, what is it? (A drawing may help)
Finally, note that $x_j = a + j\Delta x$, where $\Delta x$ is the length of any of the n uniform interval (with this substitution we end up with sums of $j$ or $j^2$ that can be reduced to expressions in $n$).
Detailed Hint. $S(n) = \sum_{j=1}^n x_j^2 \Delta x = n a^2\Delta x + 2a\Delta x^2 \sum_{j=1}^n j + \Delta x^3 \sum_{j=1}^n j^2$, but $\sum_{j=1}^n j = \frac{n(n+1)}{2}$ and $\sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$, and $\Delta x = \frac{b-a}{n}$, then we may take the limit over $n$.
A: For the mid-point case (iii), you get:
Interval: $I_j=[a+\frac{(b-a)(j-1)}{n}, \,a+\frac{(b-a)(j)}{n}]$, and thus
$$S(n)=\sum_{j = 1}^n \bigg(a+\frac{(b-a)j}{n}-\frac{b-a}{2n}\bigg)^2\cdot \frac{b-a}{n}$$
A: As you say, iii) is indeed a definition of $\epsilon_j$. 
And its value is given by a simple formula $\epsilon_j = \frac{x_{j-1}+x_j}{2}$. 
And it's something you can then evaluate the function $f$ on:
$$f(\epsilon_j) = \bigl(\frac{x_{j-1}+x_j}{2}  \bigr)^2 = \frac{x_{j-1}^2 + 2 x_{j-1} x_j + x_j^2}{4}
$$
