# How do you find the area using the trapezoidal approximation method, given (x²-6x+9) dx, on the interval [0,3] with n=3?

Oct 29, 2017

${\int}_{0}^{3} \setminus {x}^{2} - 6 x + 9 \setminus \mathrm{dx} \approx 9.5$

#### Explanation:

We have:

$y = {x}^{2} - 6 x + 9$

We want to estimate $\int \setminus y \setminus \mathrm{dx}$ over the interval $\left[0 , 3\right]$ with $3$ strips; thus:

$\Delta x = \frac{3 - 0}{3} = 1$

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

$A = {\int}_{0}^{3} \setminus {x}^{2} - 6 x + 9 \setminus \mathrm{dx}$
$\setminus \setminus \setminus \approx \frac{1}{2} \cdot \left\{9 - 0 + 2 \cdot \left(4 + 1\right)\right\}$
$\setminus \setminus \setminus = 0.5 \cdot \left\{9 + 2 \cdot \left(5\right)\right\}$
$\setminus \setminus \setminus = 0.5 \cdot \left\{9 + 10\right\}$
$\setminus \setminus \setminus = 0.5 \cdot 19$
$\setminus \setminus \setminus = 9.5$

Actual Value

For comparison of accuracy:
$A = {\int}_{0}^{3} \setminus {x}^{2} - 6 x + 9 \setminus \mathrm{dx}$

$\setminus \setminus \setminus = {\left[{x}^{3} / 3 - 3 {x}^{2} + 9 x\right]}_{0}^{3}$
$\setminus \setminus \setminus = 9 - 27 + 27$
$\setminus \setminus \setminus = 9$