Riemann Sum Calculator - Approximate Integral with Rectangles

Approximate integrals using Riemann sums with our free online Riemann sum calculator. Choose from left, right, midpoint, or trapezoidal methods with detailed step-by-step calculations.

Function f(x):

Method:

Lower Bound (a):

Upper Bound (b):

Number of Intervals (n):

Examples:

Understanding Riemann Sums

A Riemann sum approximates the area under a curve by dividing it into rectangles (or trapezoids) and adding up their areas. Named after German mathematician Bernhard Riemann, this method is the foundation of integral calculus.

The more rectangles you use (larger n), the better the approximation. As n approaches infinity, the Riemann sum converges to the exact definite integral. This beautiful connection between sums and integrals is the Fundamental Theorem of Calculus.

Riemann Sum Methods

Left Riemann Sum

Uses left endpoint of each subinterval

For increasing functions, left sums underestimate. For decreasing functions, they overestimate.

Right Riemann Sum

Uses right endpoint of each subinterval

Opposite of left sum - overestimates for increasing functions, underestimates for decreasing.

Midpoint Rule

Uses midpoint of each subinterval

Generally more accurate than left or right sums. Errors tend to cancel out.

Trapezoidal Rule

Uses trapezoids instead of rectangles

Average of left and right sums. Often more accurate, especially for smooth curves.

Worked Examples

Example 1: Left Riemann Sum for f(x) = x²

Approximate ∫x²dx from 0 to 1 using 4 rectangles (left endpoints).

Interval: [0, 1], n = 4

Width: Δx = (1-0)/4 = 0.25

Left endpoints: 0, 0.25, 0.5, 0.75

Heights: f(0)=0, f(0.25)=0.0625, f(0.5)=0.25, f(0.75)=0.5625

Areas: 0×0.25, 0.0625×0.25, 0.25×0.25, 0.5625×0.25

Sum = 0 + 0.015625 + 0.0625 + 0.140625 = 0.21875

Exact integral = 1/3 ≈ 0.333... Left sum underestimates

Example 2: Right Riemann Sum for f(x) = x²

Same function, but using right endpoints.

Right endpoints: 0.25, 0.5, 0.75, 1

Heights: f(0.25)=0.0625, f(0.5)=0.25, f(0.75)=0.5625, f(1)=1

Sum = (0.0625 + 0.25 + 0.5625 + 1) × 0.25 = 0.46875

Right sum overestimates (0.469 > 0.333)

Example 3: Trapezoidal Rule

The trapezoidal rule averages left and right sums.

Trapezoidal = (Left + Right) / 2

Trapezoidal = (0.21875 + 0.46875) / 2 = 0.34375

Much closer to exact value of 0.333!

Example 4: Effect of More Intervals

What happens when we increase n for f(x) = x² on [0,1]?

n = 4: Left = 0.21875, Right = 0.46875

n = 10: Left = 0.2850, Right = 0.3850

n = 100: Left = 0.3283, Right = 0.3383

n = 1000: Left = 0.3328, Right = 0.3338

As n increases, both approach 1/3 ≈ 0.3333...

Quick Fact

Bernhard Riemann (1826-1866) developed his theory of integration in his 1854 habilitation thesis. Tragically, he died of tuberculosis at age 39. Despite his short life, Riemann revolutionized mathematics - his name appears in the Riemann sum, Riemann integral, Riemann hypothesis (the most famous unsolved problem in mathematics), and Riemannian geometry (the foundation of Einstein's general relativity).

Frequently Asked Questions

Which Riemann sum method is most accurate?

For the same number of intervals, midpoint and trapezoidal rules are generally more accurate than left or right sums. Simpson's rule (not implemented here) is even more accurate but requires parabolic approximations instead of straight lines.

How many intervals should I use?

More intervals = better accuracy but more computation. For smooth functions, n=10 to n=100 often gives good results. For highly curved functions, you may need more. The error typically decreases proportionally to 1/n (or 1/n² for midpoint/trapezoidal).

When would left sum equal right sum?

Left and right sums are equal only when the function is constant (horizontal line). For any non-constant function, they'll differ. The difference between them equals (f(b) - f(a)) × Δx.

Can Riemann sums handle negative function values?

Yes! When f(x) is negative, the rectangle has "negative area" - it's below the x-axis. The Riemann sum calculates the net signed area: area above the axis minus area below the axis.

What's the relationship between Riemann sums and definite integrals?

The definite integral IS the limit of Riemann sums as n→∞. Formally: ∫[a,b] f(x)dx = lim(n→∞) Σf(xᵢ*)Δx. Riemann sums are how we define and compute integrals.

Where are Riemann sums used in real applications?

Anywhere you need to find area, volume, or accumulated quantities: physics (work, displacement), engineering (stress analysis), economics (consumer surplus), probability (continuous distributions), and computer graphics (rendering).

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