How do you find the area using the trapezoidal approximation method, given  sinx(dx) , on the interval [0, pi] with n=10?

Apr 25, 2018

The answer $A = 113.652$

Explanation:

-Firstly:[sketch the function f(x)=sin(x) on interval $\left[0 , \pi\right]$

-secondly:[find the width=(b-a)/n ]

$w i \mathrm{dt} h = \frac{b - a}{n} = \frac{\pi - 0}{10} = \frac{\pi}{10}$

-Thirdly: write the parts of interval

$\left[0 , \frac{\pi}{10}\right] \left[\frac{\pi}{10} , 2 \frac{\pi}{10}\right] \left[2 \frac{\pi}{10} , 3 \frac{\pi}{10}\right] \left[3 \frac{\pi}{10} , 4 \frac{\pi}{10}\right]$

$\left[4 \frac{\pi}{10} , 5 \frac{\pi}{10}\right] \ldots \ldots . . \left[9 \frac{\pi}{10} , \pi\right]$

the sketch of function after separate it by intervals

-Fourthly calculate the area by using the law:

$A = \frac{b - a}{2 n} \left[f \left({x}_{0}\right) + 2 f \left({x}_{1}\right) + 2 f \left({x}_{2}\right) + \ldots \ldots . . + 2 f \left({x}_{n - 1}\right) + f \left({x}_{n}\right)\right]$

$A = \frac{\pi}{20} \left[f \left(0\right) + 2 f \left(\frac{\pi}{10}\right) + 2 f \left(2 \frac{\pi}{10}\right) + 2 f \left(3 \frac{\pi}{10}\right) + 2 f \left(4 \frac{\pi}{10}\right) + 2 f \left(5 \frac{\pi}{10}\right) + 2 f \left(6 \frac{\pi}{10}\right) + 2 f \left(7 \frac{\pi}{10}\right) + 2 f \left(8 \frac{\pi}{10}\right) + 2 f \left(9 \frac{\pi}{10}\right) + f \left(\pi\right)\right]$

$\frac{\pi}{20} \left[0 + 0.618 + 1.176 + 1.618 + 1.902 + 2 + 1.902 + 1.618 + 1.176 + 0.618 + 0\right]$

$A = \frac{\pi}{20} \left(12.628\right) = 9 \cdot \left(12.628\right) = 113.652$

Note $\pi = 180$