# How do you find the area using the trapezoid approximation method, given 1/x^2 dx , on the interval [1,3] with n=5?

May 21, 2017

Trapezium rule gives:

${\int}_{1}^{3} \setminus \frac{1}{x} ^ 2 \setminus \mathrm{dx} \approx 0.69$ (2dp)

#### Explanation:

The values of $f \left(x\right) = \frac{1}{x} ^ 2 - 1$ 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:

${\int}_{1}^{3} \setminus \frac{1}{x} ^ 2 \setminus \mathrm{dx}$
$\text{ } \approx \frac{0.4}{2} \left\{1 + 0.11111 + 2 \left(0.5102 + 0.30864 + 0.20661 + 0.14793\right)\right\}$

$\text{ } = 0.2 \left\{1.11111 + 2 \left(1.17339\right)\right\}$
$\text{ } = 0.2 \left\{1.11111 + 2.34677\right\}$
$\text{ } = 0.2 \cdot 3.45788$
$\text{ } = 0.69158$

Let's compare this to the exact value:

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