# What are the requirements of a function so that the left Riemann sum equals the right Riemann sum?

My homework question in particular specifies over an interval of [0,1], the function is negative, and the function is decreasing.

• What is a "right" or "left" Riemann Sum?? – DonAntonio Jan 20 at 23:17
• @DonAntonio Presumably a Riemann sum in which one takes the value of the function at the right or left endpoints of the subintervals. – Math1000 Jan 21 at 0:22

Given: $$\Delta x = \frac{b - a}{n}$$, when no limits are involved we can expand the left and right Riemann sums as follows:

\begin{align*} R_L &= \sum_{k = 0}^{n -1} f(a + k\Delta x) \Delta x = f(a) +\sum_{k = 1}^{n -1} f(a + k\Delta x) \Delta x \\ R_R &= \sum_{k = 1}^{n} f(a + k\Delta x) \Delta x = f(a+ n \Delta x) + \sum_{k = 1}^{n - 1} f(a + k\Delta x) \Delta x \\ \end{align*}

If $$R_L = R_R$$, then their difference should be zero:

\begin{align*} R_L - R_R &= 0\\ \left(f(a) +\sum_{k = 1}^{n -1} f(a + k\Delta x) \Delta x\right) - \left( f(a+ n \Delta x) + \sum_{k = 1}^{n - 1} f(a + k\Delta x) \Delta x\right) &= 0 \\ f(a) - f(a+ n \Delta x) &= 0 \\ \end{align*}

Therefore, the left and right Riemann sums are the same, without limits, when the following holds true:

$$f(a) = f(a + n \Delta x)$$

When limits are involved, the left Riemann sum approaches the right Riemann sum, as $$n \to \infty$$.

\begin{align*} \lim_{n \to \infty} R_L &= \lim_{n \to \infty} R_R \\ \end{align*}

• Isn't the requirement for non-limit situations much looser, namely that the left and right ends must be equal? – Dan Uznanski Jan 25 at 9:51
• @DanUznanski I might be overlooking something. Can you explain what you mean by the left and right ends? – Gustav Jan 25 at 9:59
• $R_R - R_L = \sum_{k=1}^{n} df(a+kd) -\sum_{k=0}^{n-1} df(a+kd) = \sum_{k=1}^{n-1} df(a+kd) + df(a+nd) - (\sum_{k=1}^{n-1} df(a+kd) + df(a+0d) = d(f(a+nd)-f(a))$ so equality is when $f(a+nd)=f(a)$ – Dan Uznanski Jan 25 at 10:09
• @DanUznanski You are right. Do you want to submit this as an answer, or should I edit this into my answer? – Gustav Jan 25 at 11:33
• Think it belongs in yours – Dan Uznanski Jan 25 at 11:34