Riemann Sum — Left Endpoint Approximation

April 12, 2026

Problem

Approximate the area under y = x² from 0 to 4 using 8 left-endpoint rectangles.

Explanation

A Riemann sum approximates the area under a curve by stacking rectangles. The "left-endpoint" version uses the function value at the left side of each subinterval to determine the rectangle's height. For an increasing function, this systematically underestimates the true area.

The Setup

Divide [a,b][a, b] into nn equal subintervals of width:

Δx=ban\Delta x = \dfrac{b - a}{n}

The left endpoints are xi=a+iΔxx_i = a + i\,\Delta x for i=0,1,,n1i = 0, 1, \ldots, n - 1. The left Riemann sum is:

Ln=i=0n1f(xi)ΔxL_n = \sum_{i=0}^{n-1} f(x_i)\,\Delta x

Step-by-Step Solution

Given: f(x)=x2f(x) = x^{2}, a=0a = 0, b=4b = 4, n=8n = 8.

Find: The left-endpoint Riemann sum L8L_8.


Step 1 — Compute Δx\Delta x.

Δx=ban=408=0.5\Delta x = \dfrac{b - a}{n} = \dfrac{4 - 0}{8} = 0.5

Step 2 — List the left endpoints.

xi=0+i(0.5)    for    i=0,1,,7x_i = 0 + i(0.5) \;\;\text{for}\;\; i = 0, 1, \ldots, 7

x0,x1,,x7=0,  0.5,  1.0,  1.5,  2.0,  2.5,  3.0,  3.5x_0, x_1, \ldots, x_7 = 0,\;0.5,\;1.0,\;1.5,\;2.0,\;2.5,\;3.0,\;3.5

Step 3 — Evaluate f(xi)=xi2f(x_i) = x_i^{2} at each.

f(0)=0,    f(0.5)=0.25,    f(1)=1,    f(1.5)=2.25,f(0) = 0, \;\; f(0.5) = 0.25, \;\; f(1) = 1, \;\; f(1.5) = 2.25,

f(2)=4,    f(2.5)=6.25,    f(3)=9,    f(3.5)=12.25f(2) = 4, \;\; f(2.5) = 6.25, \;\; f(3) = 9, \;\; f(3.5) = 12.25

Step 4 — Sum the function values.

i=07f(xi)=0+0.25+1+2.25+4+6.25+9+12.25=35.0\sum_{i=0}^{7} f(x_i) = 0 + 0.25 + 1 + 2.25 + 4 + 6.25 + 9 + 12.25 = 35.0

Step 5 — Multiply by Δx\Delta x.

L8=0.5×35.0=17.5L_8 = 0.5 \times 35.0 = 17.5

Step 6 — Compare to the exact value.

04x2dx=x3304=64321.333\int_{0}^{4} x^{2}\,dx = \dfrac{x^{3}}{3}\bigg|_{0}^{4} = \dfrac{64}{3} \approx 21.333

The error is 21.33317.5=3.83321.333 - 17.5 = 3.833, or about 18% underestimation. This makes sense: f(x)=x2f(x) = x^{2} is increasing, so the left endpoint of each subinterval is the lowest value of ff on that subinterval — every rectangle sits below the curve.


Answer: The left-endpoint Riemann sum with n=8n = 8 rectangles is

  L8=17.5  \boxed{\;L_8 = 17.5\;}

This underestimates the exact area 64321.333\dfrac{64}{3} \approx 21.333 by approximately 3.83 square units (≈ 18% error). To improve the estimate, use more rectangles, switch to right or midpoint sums, or use Simpson's rule.

Try It

  • Adjust n (number of rectangles) — watch the error shrink as nn grows.
  • Toggle show exact area to see the true area highlighted in green.
  • The HUD shows the computed LnL_n alongside the exact value 64/364/3 for comparison.
  • Notice how every rectangle's top-left corner touches the curve — that's the defining feature of a left Riemann sum.

Interactive Visualization

Parameters

8.00
Your turn

Got your own math or physics problem?

Turn any problem into an interactive visualization like this one — powered by AI, generated in seconds. Free to try, no credit card required.

Sign Up Free to Try It30 free visualizations every day
Riemann Sum — Left Endpoint Approximation | MathSpin