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

Question

Christopher Allers · Accepted Answer

Trapezium rule gives:

# int_2^3 \ 1/(x-1)^2 \ dx ~~ 0.51 # (2dp)
The values of #f(x)=1/(x-1)^2# are tabulated as follows (using Excel) working to 5dp

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 #

Luke Alvarez · Answer

undefined To estimate the area under the curve $ \frac{1}{{(x-1)}^2} $ over the interval $[2,3]$ with $n=4$ using the trapezium rule, you would first divide the interval $[2,3]$ into $n$ subintervals of equal width. In this case, with $n=4$, each subinterval would have a width of $ \frac{3-2}{4} = 0.25$.

Then, apply the trapezium rule formula for each subinterval:

$$ A_i = \frac{h}{2} \left( f(x_i) + f(x_{i+1}) \right) $$

where $A_i$ is the area of the $i$th trapezium, $h$ is the width of each subinterval, and $f(x_i)$ and $f(x_{i+1})$ are the function values at the endpoints of the $i$th subinterval.

For $i = 1, 2, 3, 4$, calculate $f(x_i)$ and $f(x_{i+1})$ and substitute them into the formula. Finally, sum up the areas of all the trapeziums to get the estimated area under the curve over the interval $[2,3]$.