6.2 Approximating Areas with Riemann Sums

Syllabus
2020
Topic
6.2
Level

Learning objectives

Approximate an Integral from a Partition

To approximate abf(x)dx\int_a^b f(x)\,dx, partition [a,b][a,b] at a=x0<x1<<xn=ba=x_0<x_1<\cdots<x_n=b. On each subinterval, multiply its own width Δxi=xixi1\Delta x_i=x_i-x_{i-1} by a height chosen by the stated method, then add. This works for equal or unequal widths.

Method Height on [xi1,xi][x_{i-1},x_i] Useful error clue
Left f(xi1)f(x_{i-1}) increasing: under; decreasing: over
Right f(xi)f(x_i) increasing: over; decreasing: under
Midpoint f((xi1+xi)/2)f((x_{i-1}+x_i)/2) concave up: under; concave down: over
Trapezoidal [f(xi1)+f(xi)]/2[f(x_{i-1})+f(x_i)]/2 concave up: over; concave down: under

L=\sum_{i=1}^n f(x_{i-1})\Delta x_i,\quad R=\sum_{i=1}^n f(x_i)\Delta x_i,\quad T=\sum_{i=1}^n \frac{f(x_{i-1})+f(x_i)}{2}\Delta x_i

Suppose a table gives x=0,1,3,4x=0,1,3,4 and f(x)=2,4,5,3f(x)=2,4,5,3. The widths are 1,2,11,2,1, not all equal. The left sum is 2(1)+4(2)+5(1)=152(1)+4(2)+5(1)=15. The right sum is 4(1)+5(2)+3(1)=174(1)+5(2)+3(1)=17. The trapezoidal sum is 2+42(1)+4+52(2)+5+32(1)=16\tfrac{2+4}{2}(1)+\tfrac{4+5}{2}(2)+\tfrac{5+3}{2}(1)=16.

Do not replace every Δxi\Delta x_i by (ba)/n(b-a)/n unless the partition is uniform. An error direction also requires the relevant behavior across the interval: monotonicity supports left/right judgments, while concavity supports midpoint/trapezoidal judgments. Without that information, calculate the approximation but do not guess over or under.