Limits of Riemann Sums

With $n$ subintervals of equal width and right endpoints, the limit that defines the integral of $f$ from $a$ to $b$ is

Three things can be read off a limit written this way. The factor outside $f$ is the width $\Delta x = \frac{b-a}{n}$, and its numerator is the length of the interval. The expression inside $f$ is the right endpoint $x_i = a + i\,\Delta x$, and its constant term is the lower limit $a$. Finally, $f$ is whatever is done to $x_i$.

The choice of $x_i$ in Example 1 wasn't forced. The constant $1$ in the exponent could belong to the function instead of to the endpoint.

The usual mistake mixes the two answers. It takes the lower limit $0$ from Example 2 and the function from Example 1, which gives the region under $y = e^x$ from $0$ to $2$. Its heights run from $1$ to about $7.4$, where the correct region's run from about $2.7$ to about $20.1$, so its area is much smaller.

In Example 1 the factor outside $f$ matched the step inside it: both were $\frac{2}{n}$. When they don't match, the integral can be written two ways. One keeps the factor outside as $\Delta x$ and makes the step inside part of the function. The other rewrites the factor outside to match the step. Example 3 does both.

Dropping the $\frac{1}{2}$ is the usual mistake here. It leaves the integral of the exponential from $0$ to $2$, which is twice the correct value.

Sometimes the function is a line or a piece of a circle, and its graph stays on or above the axis. Then the integral is an area, an area formula evaluates it, and the limit needs no summation formulas at all. The properties lesson takes up graphs that go below the axis.

Going the other way takes the same three parts. Find $\Delta x$ from the length of the interval, write the right endpoint $x_i = a + i\,\Delta x$, and put $x_i$ into $f$.

Left endpoints start one step earlier. The left endpoint of the $i$-th subinterval is $x_{i-1} = a + (i - 1)\,\Delta x$, so the first one is $a$ itself and the last is $b - \Delta x$.