1. ## function integrable

Suppose $a. Let $k$ be any non-zero real number. Define $f: \mathbb{R} \to \mathbb{R}$ as follows: $f(x) = \begin{cases} 0 \ \ \ \ \ \text{if} \ x < c \\ k \ \ \ \ \ \text{if} \ c \leq x \leq d \\ 0 \ \ \ \ \ \text{if} \ d < x \end{cases}$

(a) Prove that $f$ is Riemann Integrable on $[a,b]$. What is its integral?

(b) If every $<$ in definition of $f$ were changed to $\leq$ and vice versa, how would the answer change?

For (a) the integral would be $k(d-c)$. Here is a proof. Proof. Let $\epsilon >0$. Choose $\delta >0$ such that $\delta < \frac{\epsilon}{2|d-c|}$. Let $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n-1}, x_{n} \}$ be a partition of $[a,b]$ with $||P|| < \delta$. Choose $x_{1}^{*}, x_{2}^{*}, \ldots , x_{n}^{*}$ arbitrarily such that $x_{i-1} \leq x_{i}^{*} \leq x_{i}$. Then $\mathcal{R}(f,P) = \sum_{i=1}^{n} f(x_{i}^{*})(x_{i}-x_{i-1})$. We have to break this up into three sums? Because ultimately we want to show that $|\mathcal{R}(f,P)-k(d-c)| < \epsilon$.

2. For any such partition, you can make a finer partition by adding c and d as "break points". Once you do that, the three sums are simple.

3. Originally Posted by HallsofIvy
For any such partition, you can make a finer partition by adding c and d as "break points". Once you do that, the three sums are simple.
Because then two sums are $0$ and we are only worried about the "middle" sum?