# How do you find the area using the trapezoidal approximation method, given f(x)=x^2 -1, on the interval [2,4] with n=8?

Mar 14, 2017

Trapezium rule gives:

${\int}_{2}^{4} \setminus {x}^{2} - 1 \setminus \mathrm{dx} \approx 16.69$ (2dp)

#### Explanation:

The values of $f \left(x\right) = {x}^{2} - 1$ are tabulated as follows (using Excel) working to 6dp

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:

${\int}_{2}^{4} \setminus {x}^{2} - 1 \setminus \mathrm{dx}$
 " " ~~ 0.25/2 { (3 + 15) + 2(4.0625 + 5.25 + 6.5625 +
 " " 8 + 9.5625 + 11.25 + 13.0625) }
$\text{ } = 0.125 \left\{18 + 2 \left(57.75\right)\right\}$
$\text{ } = 0.125 \left\{18 + 115.5\right\}$
$\text{ } = 0.125 \left\{133.5\right\}$
$\text{ } = 16.6875$

Let's compare this to the exact value:

${\int}_{2}^{4} \setminus {x}^{2} - 1 \setminus \mathrm{dx} = {\left[{x}^{3} / 3 - x\right]}_{2}^{4}$
$\text{ } = \left(\frac{64}{3} - 4\right) - \left(\frac{8}{3} - 2\right)$
$\text{ } = \frac{50}{3}$
$\text{ } = 16.666666 \ldots$