How do you find the area using the trapezoidal approximation method, given #(x²-6x+9) dx#, on the interval [0,3] with n=3?

1 Answer
Oct 29, 2017

Answer:

# int_0^3 \ x^2-6x+9 \ dx ~~ 9.5 #

Explanation:

We have:

# y = x^2-6x+9 #

We want to estimate #int \ y \ dx# over the interval #[0,3]# with #3# strips; thus:

# Deltax = (3-0)/3 = 1#

Note that we have a fixed interval (strictly speaking a Riemann sum can have a varying sized partition width). The values of the function are tabulated as follows;

Steve M using Microsoft Excel

Trapezium Rule

# A = int_0^3 \ x^2-6x+9 \ dx #
# \ \ \ ~~ 1/2 * { 9 - 0 + 2*(4 + 1) } #
# \ \ \ = 0.5 * { 9 + 2*(5) } #
# \ \ \ = 0.5 * { 9 + 10 } #
# \ \ \ = 0.5 * 19 #
# \ \ \ = 9.5 #

Actual Value

For comparison of accuracy:
# A = int_0^3 \ x^2-6x+9 \ dx #

# \ \ \ = [x^3/3-3x^2+9x]_0^3 #
# \ \ \ = 9-27+27 #
# \ \ \ = 9 #