A Left Riemann Sum approximates a definite integral by splitting the interval into subintervals and building a rectangle on each one whose height is the function's value at the LEFT endpoint. It underestimates increasing functions and overestimates decreasing ones (AP Calc Topic 6.2, LIM-5.A.2).
A Left Riemann Sum is one of the four approximation methods the CED names for estimating a definite integral (LIM-5.A.2), alongside right, midpoint, and trapezoidal sums. You chop the interval [a, b] into subintervals, and on each subinterval you build a rectangle. The height of each rectangle is the function's value at the left endpoint of that subinterval. Add up all the rectangle areas (height times width) and you have your approximation of .
Here's the intuition. Each rectangle 'locks in' the function's value at the start of its subinterval and pretends the function stays flat across it. If the function is climbing, that left-edge height is the lowest point on the subinterval, so every rectangle falls short and the sum underestimates the true area. If the function is falling, the left edge is the highest point, so the sum overestimates. The CED (LIM-5.A.4) expects you to make exactly this kind of over/under call. One more thing the CED is explicit about: partitions don't have to be uniform. On table-based FRQs, the subinterval widths are usually unequal, so compute each width separately instead of assuming one shared Δx.
Left Riemann Sums live in Topic 6.2 (Approximating Areas with Riemann Sums) in Unit 6, Integration and Accumulation of Change, supporting learning objective 6.2.A. They matter for two reasons. Conceptually, Riemann sums are how the definite integral is defined in Topic 6.3, so the left sum is your first concrete picture of what an integral actually is, infinitely many skinny rectangles. Practically, this is one of the most reliably tested skills on the AP exam. When a function is given as a table of values (LIM-5.A.1 says integrals can be approximated from numerical data), you can't antidifferentiate anything. A left Riemann sum is the tool that turns those table values into an estimate of total accumulation, like total gallons pumped or total bacteria in a dish.
Keep studying AP Calculus Unit 6
Visual cheatsheet
view galleryRight Riemann Sum (Unit 6)
The mirror image. A right sum uses the right endpoint of each subinterval for the height, so its over/under behavior flips. For an increasing function, left underestimates and right overestimates, and the true integral sits between them.
Trapezoidal Rule (Unit 6)
A trapezoidal sum is literally the average of the left and right Riemann sums. It replaces flat rectangle tops with slanted ones, which is why it usually lands closer to the true value than either endpoint sum alone.
The Definite Integral as a Limit (Unit 6, Topic 6.3)
The definite integral is defined as the limit of Riemann sums as the number of subintervals goes to infinity. The left sum isn't just an approximation trick; it's a snapshot of what an integral is before the rectangles get infinitely thin.
Accumulation and Rates of Change (Units 6 and 8)
On the exam, left sums almost always show up attached to a real rate, like gallons per second or bacteria density. The sum approximates total accumulation, which connects directly to the average value and applied integration problems later in the course.
This is a workhorse on both sections. MCQs ask things like 'how are the heights determined in a left Riemann sum?' or 'which sum always overestimates a decreasing function?' so you need the endpoint rule and the over/under logic cold. On FRQs, the classic setup is a table of values for a continuous function with a prompt like 'use a left Riemann sum with the subintervals indicated by the table to approximate the integral.' The 2021 FRQ (bacteria density in a petri dish) did exactly this, and the 2017 tank problem, 2022 melting cone, and 2023 gas pump problems all used table-based Riemann sum approximations. Two things earn the points. First, show the sum explicitly, each value times each width, with the widths read off the table (they're usually unequal). Second, be ready for the follow-up question asking whether your answer over- or underestimates the integral, justified by saying the function is increasing or decreasing on the interval. Also interpret your number with units, since these problems are always about a real quantity accumulating.
Both methods build rectangles on subintervals; the only difference is which endpoint sets the height. Left sums sample f at the start of each subinterval, right sums at the end. The classic trap is mixing up the estimate directions. For an INCREASING function, the left sum is an UNDERestimate (it grabs the smallest value on each piece) and the right sum is an OVERestimate. For a decreasing function, both flip. Quick check from a table: a left sum uses every value except the last one, a right sum uses every value except the first.
A Left Riemann Sum approximates a definite integral by using the function's value at the left endpoint of each subinterval as the rectangle height.
For an increasing function, a left Riemann sum always underestimates the integral; for a decreasing function, it always overestimates.
Subintervals do not have to be equal width, and on table-based FRQs you should compute each width separately from the table values.
When justifying over- or underestimate on an FRQ, state whether the function is increasing or decreasing on the interval; that's the reasoning that earns the point.
From a table with n+1 values, a left Riemann sum with n subintervals uses every value except the last one.
Left Riemann Sums are the foundation for the definition of the definite integral as a limit of sums in Topic 6.3.
It's a method for approximating a definite integral by dividing the interval into subintervals and using the function's value at the left endpoint of each subinterval as the rectangle height. It's one of four approximation methods named in the CED under Topic 6.2 (LIM-5.A.2).
No. It underestimates only when the function is increasing on the interval. If the function is decreasing, the left endpoint is the highest point of each subinterval, so the left sum overestimates. You have to check the function's behavior before making the call.
A left sum uses the left endpoint of each subinterval for the height; a right sum uses the right endpoint. Their estimate directions are opposite, so for an increasing function the left sum underestimates while the right sum overestimates.
Multiply each function value (skipping the last one in the table) by the width of the subinterval that starts at that point, then add the products. Watch for unequal widths, which are common on FRQs like the 2021 bacteria problem, and show the full sum for the point.
No. The CED (LIM-5.A.2) explicitly allows nonuniform partitions, and most table-based FRQs use unequal subintervals on purpose. Read each width directly from the table rather than assuming one fixed Δx.