# How do you find the area using the trapezoid approximation method, given sin (x^2) dx, on the interval [0, 1/2] using n=4?

Dec 15, 2016

${\int}_{0}^{\frac{1}{2}} \sin \left({x}^{2}\right) \mathrm{dx} \approx 0.04274 \text{ } \left(5 \mathrm{dp}\right)$

#### Explanation:

The values of $y = \sin \left({x}^{2}\right)$ 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:

${\int}_{0}^{\frac{1}{2}} \sin \left({x}^{2}\right) \mathrm{dx} \approx \frac{0.125}{2} \left\{0 + 0.2474 + 2 \left(0.01562 + 0.06246 + 0.14016\right)\right\}$
$\text{ } = 0.0625 \left\{0.2474 + 2 \left(0.21825\right)\right\}$
$\text{ } = 0.0625 \left\{0.2474 + 0.43649\right\}$
$\text{ } = 0.0625 \left\{0.6839\right\}$
$\text{ } = 0.04274$