# How do you use the trapezoidal rule with n=9 to approximate the area between the curve y=x^2 -2x +2 from 0 to 3?

Sep 14, 2015

We use the formula $\frac{1}{6} \left[f \left({x}_{0}\right) + 2 f \left({x}_{1}\right) + 2 f \left({x}_{2}\right) + \cdots + 2 f \left({x}_{8}\right) + f \left({x}_{9}\right)\right]$ where $f \left(x\right) = {x}^{2} - 2 x + 2$ and ${x}_{0} \cdots {x}_{9}$ are chosen to evenly divide the interval $\left[0 , 3\right]$ into 9 pieces.

#### Explanation:

The trapezoidal rule works by dividing the area under the curve into $n$ trapezoids, then calculating the area of each trapezoid, and summing them up. The formula is:

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = \frac{b - a}{2 n} \left[f \left({x}_{0}\right) + 2 f \left({x}_{1}\right) + 2 f \left({x}_{2}\right) + \cdots 2 f \left({x}_{n - 1}\right) + f \left({x}_{n}\right)\right]$

In our case, $n = 9$, $a = 0$, and $b = 3$, so we have:

${\int}_{0}^{3} f \left(x\right) \mathrm{dx} = \frac{3 - 0}{2 \cdot 9} \left[f \left({x}_{0}\right) + 2 f \left({x}_{1}\right) + 2 f \left({x}_{2}\right) + 2 f \left({x}_{3}\right) + 2 f \left({x}_{4}\right) + 2 f \left({x}_{5}\right) + 2 f \left({x}_{6}\right) + 2 f \left({x}_{7}\right) + 2 f \left({x}_{8}\right) + f \left({x}_{9}\right)\right]$

We choose the ${x}_{0} \cdots {x}_{9}$ so that they divide up the interval from 0 to 3 evenly. So ${x}_{0} = 0 , {x}_{1} = \frac{1}{3} , {x}_{2} = \frac{2}{3} , {x}_{3} = 1 , {x}_{4} = \frac{4}{3} , {x}_{5} = \frac{5}{3} , {x}_{6} = 2 , {x}_{7} = \frac{7}{3} , {x}_{8} = \frac{8}{3} , {x}_{9} = 9$.

From there it's just a matter of plugging these numbers into $f \left(x\right)$ and solving the formula.