# How do you use the trapezoidal rule with n = 4 to approximate the area bounded by the curves y = sin2x, the lines x = 0, y = 0, and x = 1?

Nov 3, 2016

${\int}_{0}^{1} \sin \left(2 x\right) \mathrm{dx} \approx 0.693$ (3dp)

#### Explanation:

The values of y are tabulated as follows (using Excel)

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_0^1sin(2x)dx ~~ 0.25/2{ (0.00000 + 0.90930 + 2(0.47943 + 0.84147 + 0.99749)}

$\therefore {\int}_{0}^{1} \sin \left(2 x\right) \mathrm{dx} \approx 0.125 \left\{0.90930 + 2 \left(2.31839\right)\right\}$
$\therefore {\int}_{0}^{1} \sin \left(2 x\right) \mathrm{dx} \approx 0.125 \left\{0.90930 + 4.63678\right\}$
$\therefore {\int}_{0}^{1} \sin \left(2 x\right) \mathrm{dx} \approx 0.125 \left\{5.54608\right\}$
$\therefore {\int}_{0}^{1} \sin \left(2 x\right) \mathrm{dx} \approx 0.69326$