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

Mar 13, 2017

Trapezium rule gives:

${\int}_{2}^{3} \setminus \frac{1}{x - 1} ^ 2 \setminus \mathrm{dx} \approx 0.51$ (2dp)

#### Explanation:

The values of $f \left(x\right) = \frac{1}{x - 1} ^ 2$ are tabulated as follows (using Excel) working to 5dp

Using the trapezoidal rule:

${\int}_{a}^{b} y \mathrm{dx} \approx \frac{h}{2} \left\{\left({y}_{0} + {y}_{n}\right) + 2 \left({y}_{1} + {y}_{2} + \ldots + {y}_{n - 1}\right)\right\}$

We have:

${\int}_{2}^{3} \setminus \frac{1}{x - 1} ^ 2 \setminus \mathrm{dx}$
$\text{ } \approx \frac{0.25}{2} \left\{\left(1 + 0.25\right) + 2 \left(0.64 + 0.444444 + 0.32653\right)\right\}$
$\text{ } = 0.125 \left\{1.25 + 2 \left(1.410975\right)\right\}$
$\text{ } = 0.125 \left\{1.25 + 2.82195\right\}$
$\text{ } = 0.125 \left\{4.07195\right\}$
$\text{ } = 0.508993$

Let's compare this to the exact value:

${\int}_{2}^{3} \setminus \frac{1}{x - 1} ^ 2 \setminus \mathrm{dx} = {\left[- \frac{1}{x - 1}\right]}_{2}^{3}$
$\text{ } = - \left(\frac{1}{2} - 1\right)$
$\text{ } = 0.5$