Riemann Sum: Definition, Formula, and Example

A Riemann sum approximates a definite integral with a single repeated calculation: width times height on each slice, then add the slices. Get the width and the sample point straight and the rest is arithmetic.

The formula and its symbols

For a function $f$ on $[a,b]$, the general Riemann sum is

Here $\Delta x_i$ is the width of the $i$th subinterval, and $x_i^*$ is the sample point chosen inside that subinterval — its height. The sample point can be the left endpoint, the right endpoint, or the midpoint. Each term $f(x_i^*)\,\Delta x_i$ is one small contribution; the sum adds them across the interval.

Why this approximates the integral

The definite integral $\int_a^b f(x)\,dx$ measures accumulated change over an interval and often represents signed area under the curve. A Riemann sum replaces the curve with rectangles that are easy to add. As the slices get thinner, the rectangles follow the graph more closely — and if $f$ is continuous on $[a,b]$, finer partitions make the sums approach the exact integral. That limiting process is the core idea behind integration: the integral is the limit of these small accumulated pieces. The formula simply makes the "small contribution times width, then add" structure visible before limit notation hides it.

Left, right, and midpoint sums

With equal-width subintervals,

and the sample-point rule sets the type:

• Left sum: left endpoint of each subinterval.
• Right sum: right endpoint of each subinterval.
• Midpoint sum: midpoint of each subinterval.

These choices can change whether the estimate is too high or too low. If the function is increasing on the whole interval, a left sum underestimates and a right sum overestimates — a conclusion that depends on the function actually being increasing there.

Worked example: right sum for $f(x)=x^2$ on $[0,2]$

Approximate $\int_0^2 x^2\,dx$ with a right Riemann sum, $n=4$ equal subintervals.

Width of each subinterval:

Right endpoints:

Evaluate $f$:

Build the sum:

The exact integral is

The comparison matters more than the arithmetic: the right sum gives $3.75$ while the exact value is about $2.67$, so the right sum overestimates. That happens because $x^2$ is increasing on $[0,2]$, making each right-endpoint rectangle slightly too tall.

Practice the calculation

Compute a midpoint sum for $f(x)=x^2$ on $[0,2]$ with the same $n=4$. The midpoints are $\tfrac{1}{4}, \tfrac{3}{4}, \tfrac{5}{4}, \tfrac{7}{4}$. Answer check: the midpoint estimate should come out to $\tfrac{21}{8} = 2.625$, which lands much closer to the exact $\tfrac{8}{3} \approx 2.67$ than the right sum did — a quick way to see how the sample-point choice changes the estimate.

Calculation traps to watch for

1. Mixing up $\Delta x$ and the sample point. $\Delta x$ is the width; $x_i^*$ is where you measure the height.
2. Using right endpoints when the problem asks for left endpoints, or vice versa.
3. Forgetting the estimate can be above or below the exact integral depending on the function and the sample rule.
4. Treating every Riemann sum as ordinary area. If the function is below the $x$-axis, the sum gives signed area, not total physical area.
5. Assuming more rectangles always fix everything. Finer partitions improve the approximation under standard conditions such as continuity, but the sum is still an approximation until you take the limit.

When Riemann sums are used

Riemann sums show up wherever a quantity is built from many small contributions: in calculus they give the intuition behind the definite integral; in physics they model accumulated quantities such as displacement from velocity samples; in numerical work they give simple estimates when an exact antiderivative is inconvenient. They are also a practical check on whether you understand what an integral means before applying antiderivative rules mechanically.