# How do you find the area using the trapezoidal approximation method, given cos(4 x) dx, on the interval [-1, 2] with n=10?

Sep 19, 2017

${\int}_{- 1}^{2} \setminus \cos 4 x \setminus \mathrm{dx} \approx 0.0510$ 4dp

#### Explanation:

We have:

$y = \cos \left(4 x\right)$

We want to estimate $\int \setminus y \setminus \mathrm{dx}$ over the interval $\left[- 1 , 2\right]$ with $n = 10$ strips; thus:

$\Delta x = \frac{2 - \left(- 1\right)}{10} = 0.3$

The values of the function are tabulated as follows;

Trapezium Rule

 A = 0.3/2 * { -0.653644 - 0.1455 +
 \ \ \ \ \ \ \ \ \ 2*(-0.942222 - 0.0292 + 0.921061 +
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus 0.696707 - 0.416147 - 0.998295 -$
 \ \ \ \ \ \ \ \ \ 0.307333 + 0.775566 + 0.869397) }

$\setminus \setminus \setminus = 0.15 \cdot \left\{- 0.799144 + 2 \cdot \left(0.569535\right)\right\}$
$\setminus \setminus \setminus = 0.15 \cdot \left\{- 0.799144 + 1.139069\right\}$
$\setminus \setminus \setminus = 0.15 \cdot 0.339926$
$\setminus \setminus \setminus = 0.050989$

Actual Value

For comparison of accuracy:

$A = {\int}_{- 1}^{2} \setminus \cos 4 x \setminus \mathrm{dx}$
 " \ \ \ = [1/4sin4x]_(-1)^2
 " \ \ \ = 1/4(sin8-sin(-4)) \ \ \ \  (radians!)
 " \ \ \ = 1/4(0.989358... - 0.756802 ...)
 " \ \ \ = 0.058138 ...