# How do you find the area using the trapezoidal approximation method, given cos(x^2), on the interval [0, 1] with n=8?

Dec 18, 2016

and so we can conclude that ${\int}_{0}^{1} \cos \left({x}^{2}\right) \mathrm{dx} \approx 0.90233$ to 5dp

#### Explanation:

The values of $y = \cos \left({x}^{2}\right)$ 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_(0)^(1) cos(x^2)dx ~~ 0.125/2 { 1 + 0.5403
 " " + 2(0.99988 + 0.99805 + 0.99013
$\text{ } + 0.96891 + 0.92467 + 0.84592$
 " " + 0.72095)}
$\text{ } = 0.0625 \left\{1.5403 + 2 \left(6.44851\right)\right\}$
$\text{ } = 0.0625 \left\{1.5403 + 12.89702\right\}$
$\text{ } = 0.0625 \left\{14.43733\right\}$
$\text{ } = 0.90233$

and so we can conclude that $y = \cos \left({x}^{2}\right) = 0.90233$ to 5dp#