How do you use the trapezoidal rule with n=4 to approximate the area between the curve #y=1/(x-1)^2# from 2 to 3?

1 Answer
Jul 17, 2017

Answer:

# int_2^3 \ 1/(x-1)^2 \ dx ~~ 0.509 #

Explanation:

We have:

# y = 1/(x-1)^2 #

We want to estimate #int \ y \ dx# over the interval #[2,3]# with #4# strips; thus:

# Deltax = (3-2)/3 = 1/4#

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;

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Trapezium Rule

# A = int_2^3 \ 1/(x-1)^2 \ dx #
# \ \ \ ~~ 0.25/2 * { 1 + 0.25 + 2*(0.64 + 0.4444 + 0.3265) } #
# \ \ \ = 0.125 * { 1.25 + 2*(1.411) } #
# \ \ \ = 0.125 * { 1.25 + 2.822 } #
# \ \ \ = 0.125 * 4.072 #
# \ \ \ = 0.509 #

Actual Value

For comparison of accuracy:

# A= int_2^3 \ 1/(x-1)^2 \ dx #
# " \ \ \ = [-1/(x-1)]_2^3 #
# " \ \ \ = [1/(x-1)]_3^2 #
# " \ \ \ = 1/1-1/2 #
# " \ \ \ = 0.5 #