Blog · Calculus guides · 27 October 2026 · 4 min read

Riemann Sums: Left, Right, Midpoint & Practice Problems

Q: Which Riemann sum is most accurate?

Usually the midpoint sum. Its error shrinks like 1/(n^2), while left and right sums shrink like 1/n, so doubling n cuts the midpoint error by about four.

A Riemann sum estimates the area under a curve by cutting $[a, b]$ into $n$ strips of width $Δ x = \frac{b - a}{n}$ , building a rectangle on each strip and adding up the areas. The left, right and midpoint sums differ only in where each rectangle takes its height: the left edge, the right edge or the middle of its strip. As $n$ grows, every version closes in on the definite integral $\int_{a}^{b} f (x) d x$ .

The formula

i = 1 \sum n f (x_{i}^{*}) Δ x, Δ x = \frac{b - a}{n}, x_{i}^{*} = ⎩ ⎨ ⎧ a + (i - 1) Δ x a + i Δ x a + (i - \frac{1}{2}) Δ x left right midpoint

Rectangles all the way down. The only decision you ever make is where on the strip the rectangle gets measured.

When to use it

Use a Riemann sum when a question asks you to estimate an integral with a fixed number of rectangles, when you only have a table of values instead of a formula, or when you want to see why the integral means area at all. If the question just says "evaluate the integral" and hands you a function you can antidifferentiate, skip the rectangles and use the Fundamental Theorem of Calculus: it gives the exact answer that every Riemann sum is chasing.

How to do it, step by step

Find the width. $Δ x = \frac{b - a}{n}$ . Every strip is the same width, so you only do this once.
List the sample points. Write out the $n$ points where you will measure the height: left edges start at $a$ , right edges end at $b$ , midpoints sit half a strip in from each.
Evaluate $f$ at each one. A small table of $x$ and $f (x)$ keeps this tidy and stops you skipping one.
Add the heights, then multiply by $Δ x$ . Factoring $Δ x$ out saves $n - 1$ multiplications and a lot of arithmetic mistakes.

Worked examples

Example 1: A left sum under a parabola

Easy

Estimate the area under $f (x) = x^{2}$ on $[0, 2]$ with a left Riemann sum and $n = 4$ .

f (x) = x^{2}, [0, 2], n = 4

The width is $Δ x = \frac{2 - 0}{4} = \frac{1}{2}$ , so the strips are $[0, 0.5]$ , $[0.5, 1]$ , $[1, 1.5]$ and $[1.5, 2]$ . A left sum measures each rectangle at its left edge: $0, 0.5, 1, 1.5$ .

L_{4} = \frac{1}{2} (0^{2} + 0. 5^{2} + 1^{2} + 1. 5^{2}) = \frac{1}{2} (0 + 0.25 + 1 + 2.25) = \frac{1}{2} (3.5)

The exact area is $\frac{8}{3} \approx 2.67$ , so this is well under. That is no accident: $x^{2}$ is increasing here, so every left edge is the lowest point of its strip and every rectangle sits below the curve.

Answer

L_{4} = \frac{7}{4}

Example 2: A right sum for a decreasing function

Medium

Estimate $\int_{1}^{3} \frac{1}{x} d x$ with a right Riemann sum and $n = 4$ .

f (x) = \frac{1}{x}, [1, 3], n = 4

Here $Δ x = \frac{3 - 1}{4} = \frac{1}{2}$ , and the right edges are $1.5, 2, 2.5, 3$ . Note the first one is not $1$ : a right sum never uses the left end of the interval.

R_{4} = \frac{1}{2} (\frac{1}{1.5} + \frac{1}{2} + \frac{1}{2.5} + \frac{1}{3}) = \frac{1}{2} (\frac{2}{3} + \frac{1}{2} + \frac{2}{5} + \frac{1}{3}) = \frac{1}{2} \cdot \frac{19}{10}

The exact value is $ln 3 \approx 1.099$ . This time the right sum is the underestimate, because $\frac{1}{x}$ is decreasing, so the right edge is the low point of each strip. Left and right swap roles depending on which way the curve slopes.

Answer

R_{4} = \frac{19}{20}

Example 3: A midpoint sum with trig

Harder

Estimate the area under one arch of $f (x) = sin x$ on $[0, π]$ with a midpoint sum and $n = 4$ .

f (x) = sin x, [0, π], n = 4

The width is $Δ x = \frac{π}{4}$ . The midpoints of the four strips are $\frac{π}{8}, \frac{3 π}{8}, \frac{5 π}{8}, \frac{7 π}{8}$ .

M_{4} = \frac{π}{4} (sin \frac{π}{8} + sin \frac{3 π}{8} + sin \frac{5 π}{8} + sin \frac{7 π}{8})

The sine curve is symmetric about $\frac{π}{2}$ , so $sin \frac{7 π}{8} = sin \frac{π}{8}$ and $sin \frac{5 π}{8} = sin \frac{3 π}{8}$ . That pairs the four terms into two, and the $\frac{π}{4}$ doubles to $\frac{π}{2}$ .

Numerically that is about $2.052$ . The exact area is $2$ , so with just four rectangles the midpoint sum is off by under 3%. Midpoints are usually the best of the three, because each rectangle overshoots on one half of its strip and undershoots on the other.

Answer

M_{4} = \frac{π}{2} (sin (\frac{π}{8}) + sin (\frac{3 π}{8})) \approx 2.052

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Find the right Riemann sum with $n = 4$ :

f (x) = x^{2} + 1, [0, 2], n = 4

Problem 2

Medium

Find the left Riemann sum with $n = 4$ :

f (x) = x, [0, 4], n = 4

Problem 3

Harder

Find the midpoint Riemann sum with $n = 4$ :

f (x) = x^{3}, [0, 2], n = 4

Common mistakes

Using $n + 1$ points. Four rectangles need four heights. If your list has five, you have included both ends, which is a left sum and a right sum glued together.
Forgetting to multiply by $Δ x$ . Adding the heights gives you a total height, not an area. The width has to come in at the end.
Mixing up which sum overestimates. For an increasing function, left sums are low and right sums are high. For a decreasing one it flips. Sketch the curve and one rectangle before you claim either.
Midpoint of the wrong thing. The midpoint rule uses the middle of each strip's $x$ -interval, not the average of the two heights. Averaging the heights is the trapezoid rule, which is a different estimate.

Practice it now

Five minutes, no account

Asymptote has no Riemann-sum problems: nobody wants to type sixteen rectangles into a timer. What it does have is the thing they converge to, definite integrals, which turn up in a Sprint from about the third problem.

Play a Sprint See it on the formula sheet

FAQ

What is the difference between left, right and midpoint Riemann sums?: Only where each rectangle's height is measured: at the left end of its strip, the right end, or the middle. The widths, the number of rectangles and the adding up are identical.
Is a Riemann sum the same as a definite integral?: Not quite. A Riemann sum is an estimate with a fixed number of rectangles. The definite integral is the limit of those sums as $n \to \infty$ and the strips get infinitely thin.
Which Riemann sum is most accurate?: Usually the midpoint sum. Its error shrinks like $\frac{1}{n ^{2}}$ , while left and right sums shrink like $\frac{1}{n}$ , so doubling $n$ cuts the midpoint error by about four.
How do I know if a Riemann sum is an overestimate?: For left and right sums, look at whether the function is increasing or decreasing on the interval. For midpoint sums, look at concavity instead: a concave-down curve like one arch of $sin x$ gives an overestimate.
Where can I find Riemann sums practice problems?: Right on this page: "Your turn" has three Riemann sums practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

Riemann Sums: Left, Right, Midpoint & Practice Problems

When to use it

How to do it, step by step

Find the width.

Δ x = \frac{b - a}{n}

. Every strip is the same width, so you only do this once.

List the sample points. Write out the

n

points where you will measure the height: left edges start at

a

, right edges end at

b

, midpoints sit half a strip in from each.

Evaluate $f$ at each one. A small table of

x

and

f (x)

keeps this tidy and stops you skipping one.

Add the heights, then multiply by $Δ x$ . Factoring

Δ x

out saves

n - 1

multiplications and a lot of arithmetic mistakes.

Worked examples

Example 1: A left sum under a parabola

Easy

Estimate the area under $f (x) = x^{2}$ on $[0, 2]$ with a left Riemann sum and $n = 4$ .

f (x) = x^{2}, [0, 2], n = 4

The width is $Δ x = \frac{2 - 0}{4} = \frac{1}{2}$ , so the strips are $[0, 0.5]$ , $[0.5, 1]$ , $[1, 1.5]$ and $[1.5, 2]$ . A left sum measures each rectangle at its left edge: $0, 0.5, 1, 1.5$ .

L_{4} = \frac{1}{2} (0^{2} + 0. 5^{2} + 1^{2} + 1. 5^{2}) = \frac{1}{2} (0 + 0.25 + 1 + 2.25) = \frac{1}{2} (3.5)

Answer

L_{4} = \frac{7}{4}

Example 2: A right sum for a decreasing function

Medium

Estimate $\int_{1}^{3} \frac{1}{x} d x$ with a right Riemann sum and $n = 4$ .

f (x) = \frac{1}{x}, [1, 3], n = 4

Here $Δ x = \frac{3 - 1}{4} = \frac{1}{2}$ , and the right edges are $1.5, 2, 2.5, 3$ . Note the first one is not $1$ : a right sum never uses the left end of the interval.

R_{4} = \frac{1}{2} (\frac{1}{1.5} + \frac{1}{2} + \frac{1}{2.5} + \frac{1}{3}) = \frac{1}{2} (\frac{2}{3} + \frac{1}{2} + \frac{2}{5} + \frac{1}{3}) = \frac{1}{2} \cdot \frac{19}{10}

Answer

R_{4} = \frac{19}{20}

Example 3: A midpoint sum with trig

Harder

Estimate the area under one arch of $f (x) = sin x$ on $[0, π]$ with a midpoint sum and $n = 4$ .

f (x) = sin x, [0, π], n = 4

The width is $Δ x = \frac{π}{4}$ . The midpoints of the four strips are $\frac{π}{8}, \frac{3 π}{8}, \frac{5 π}{8}, \frac{7 π}{8}$ .

M_{4} = \frac{π}{4} (sin \frac{π}{8} + sin \frac{3 π}{8} + sin \frac{5 π}{8} + sin \frac{7 π}{8})

Answer

M_{4} = \frac{π}{2} (sin (\frac{π}{8}) + sin (\frac{3 π}{8})) \approx 2.052

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Find the right Riemann sum with $n = 4$ :

f (x) = x^{2} + 1, [0, 2], n = 4

Problem 2

Medium

Find the left Riemann sum with $n = 4$ :

f (x) = x, [0, 4], n = 4

Problem 3

Harder

Find the midpoint Riemann sum with $n = 4$ :

f (x) = x^{3}, [0, 2], n = 4

Common mistakes

Using $n + 1$ points. Four rectangles need four heights. If your list has five, you have included both ends, which is a left sum and a right sum glued together.

Forgetting to multiply by $Δ x$ . Adding the heights gives you a total height, not an area. The width has to come in at the end.

Mixing up which sum overestimates. For an increasing function, left sums are low and right sums are high. For a decreasing one it flips. Sketch the curve and one rectangle before you claim either.

Midpoint of the wrong thing. The midpoint rule uses the middle of each strip's

x

-interval, not the average of the two heights. Averaging the heights is the trapezoid rule, which is a different estimate.

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

What is the difference between left, right and midpoint Riemann sums?

Only where each rectangle's height is measured: at the left end of its strip, the right end, or the middle. The widths, the number of rectangles and the adding up are identical.

Is a Riemann sum the same as a definite integral?

Not quite. A Riemann sum is an estimate with a fixed number of rectangles. The definite integral is the limit of those sums as

n \to \infty

and the strips get infinitely thin.

Which Riemann sum is most accurate?

Usually the midpoint sum. Its error shrinks like

\frac{1}{n ^{2}}

, while left and right sums shrink like

\frac{1}{n}

, so doubling

n

cuts the midpoint error by about four.

How do I know if a Riemann sum is an overestimate?

For left and right sums, look at whether the function is increasing or decreasing on the interval. For midpoint sums, look at concavity instead: a concave-down curve like one arch of

sin x

gives an overestimate.

Where can I find Riemann sums practice problems?

Right on this page: "Your turn" has three Riemann sums practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

Riemann Sums: Left, Right, Midpoint & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A left sum under a parabola

Example 2: A right sum for a decreasing function

Example 3: A midpoint sum with trig

Your turn

Common mistakes

FAQ

Keep going

Riemann Sums: Left, Right, Midpoint & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A left sum under a parabola

Example 2: A right sum for a decreasing function

Example 3: A midpoint sum with trig

Your turn

Common mistakes

FAQ

Keep going