# 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?

Jul 17, 2017

${\int}_{2}^{3} \setminus \frac{1}{x - 1} ^ 2 \setminus \mathrm{dx} \approx 0.509$

#### Explanation:

We have:

$y = \frac{1}{x - 1} ^ 2$

We want to estimate $\int \setminus y \setminus \mathrm{dx}$ over the interval $\left[2 , 3\right]$ with $4$ strips; thus:

$\Delta x = \frac{3 - 2}{3} = \frac{1}{4}$

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;

Trapezium Rule

$A = {\int}_{2}^{3} \setminus \frac{1}{x - 1} ^ 2 \setminus \mathrm{dx}$
$\setminus \setminus \setminus \approx \frac{0.25}{2} \cdot \left\{1 + 0.25 + 2 \cdot \left(0.64 + 0.4444 + 0.3265\right)\right\}$
$\setminus \setminus \setminus = 0.125 \cdot \left\{1.25 + 2 \cdot \left(1.411\right)\right\}$
$\setminus \setminus \setminus = 0.125 \cdot \left\{1.25 + 2.822\right\}$
$\setminus \setminus \setminus = 0.125 \cdot 4.072$
$\setminus \setminus \setminus = 0.509$

Actual Value

For comparison of accuracy:

$A = {\int}_{2}^{3} \setminus \frac{1}{x - 1} ^ 2 \setminus \mathrm{dx}$
 " \ \ \ = [-1/(x-1)]_2^3
 " \ \ \ = [1/(x-1)]_3^2
 " \ \ \ = 1/1-1/2
 " \ \ \ = 0.5