Right Riemann Sum Calculator

Use the right edge of each slice for the height. Enter f(x), [a, b] and n to get Rₙ, every rectangle drawn and the table of right endpoints.

Reads as x2x^{2}
Try:

Right sum with n = 6

11.375

Exact ∫03x2 dx=9\int_{0}^{3} x^{2}\,dx = 9 · error 2.375

0.511.522.5324681012

The formula

R6=∑i=16f ⁣(xi)Δx,Δx=3−06=0.5R_{6} = \sum_{i=1}^{6} f\!\left(x_i\right)\Delta x,\quad \Delta x = \frac{3 - 0}{6} = 0.5
Each slice
iisample xxheightarea
00.50.250.125
1110.5
21.52.251.125
3242
42.56.253.125
5394.5

All five methods, n = 6

MethodEstimateError
Left6.875-2.13
Right11.3752.38
Midpoint8.9375-0.0625
Trapezoid9.1250.125
Simpson90

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The right riemann sum formula

Rn=Δx[f(x1)+f(x2)+⋯+f(xn)]R_n = \Delta x\left[f(x_1) + f(x_2) + \cdots + f(x_n)\right]

The right sum uses x1,x2,…,xnx_1, x_2, \dots, x_n. It never uses the left end aa.

If ff is increasing, RnR_n overestimates; if decreasing, it underestimates. The gap between them is Rn−Ln=Δx (f(b)−f(a))R_n - L_n = \Delta x\,(f(b) - f(a)).

In sigma notation with xi=a+iΔxx_i = a + i\Delta x: Rn=∑i=1nf ⁣(a+ib−an)b−anR_n = \sum_{i=1}^{n} f\!\left(a + i\tfrac{b-a}{n}\right)\tfrac{b-a}{n}, and its limit as n→∞n \to \infty is the definite integral.

Compare all five methods side by side on the Riemann sum calculator, or try another single rule: Trapezoidal rule, Simpson’s rule, Midpoint rule, Left Riemann sum.

Questions students ask

What is a right Riemann sum?

An area estimate using rectangles whose heights are the function values at the right endpoint of each subinterval: Rn=Δx∑i=1nf(xi)R_n = \Delta x\sum_{i=1}^{n} f(x_i).

Is a right Riemann sum an over- or underestimate?

An overestimate for increasing functions and an underestimate for decreasing functions.

How is the right Riemann sum used to define the integral?

Many textbooks define ∫abf(x) dx=lim⁡n→∞Rn\int_a^b f(x)\,dx = \lim_{n\to\infty} R_n. Write RnR_n with xi=a+ib−anx_i = a + i\frac{b-a}{n}, simplify with summation formulas, then take the limit.