Estimate the area under the curve #1/(x-1)^2# over the interval #[2,3]# with #n=4# using the trapezium rule?

1 Answer
Mar 13, 2017

Answer:

Trapezium rule gives:

# int_2^3 \ 1/(x-1)^2 \ dx ~~ 0.51 # (2dp)

Explanation:

The values of #f(x)=1/(x-1)^2# are tabulated as follows (using Excel) working to 5dp

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Using the trapezoidal rule:

# int_a^bydx ~~ h/2{(y_0+y_n)+2(y_1+y_2+...+y_(n-1))}#

We have:

# int_2^3 \ 1/(x-1)^2 \ dx #
# " " ~~ 0.25/2 { (1 + 0.25) + 2(0.64 + 0.444444 + 0.32653) } #
# " " = 0.125 { 1.25 + 2(1.410975) } #
# " " = 0.125 { 1.25 + 2.82195 } #
# " " = 0.125 { 4.07195 } #
# " " = 0.508993 #

Let's compare this to the exact value:

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