simpsons rule... (please let me know if I've copied this wrong!!)
w/3 [(y0+yn+4(y1+y3+y5....)+2(y2+y4+y6...)]
I got my q
∫(below)0 (above)2π /3 sin x dx (4 strips)
can I please have some help??
$\displaystyle \int_0^{\frac{2\pi}{3}} \sin{x} \, dx = \frac{\pi}{18}\left[\sin(0) + 4\sin\left(\frac{\pi}{6}\right) + 2\sin\left(\frac{\pi}{3}\right) + 4\sin\left(\frac{\pi}{2}\right) + \sin\left(\frac{2\pi}{3}\right)\right]
$
okay see I'm getting...
$\displaystyle \int_0^{\frac{2\pi}{3}} \sin{x} \, dx = pi/18 [sin (0) + sin (120) + 4 (0.5 + 1) + 2(sin 120)]
$
which goes as pi/18 [0 + root3/2 + 4(1.5) + root3]
in which I get the answer as approximate to 1.5
The answer in the textbook is 3.0
I'm not sure I completely understand your working skeeter...
$\displaystyle \frac{\Delta x}{3} \left(y_0 + 4y_1 + 2y_2 + 4y_3 + y_4\right)$
$\displaystyle \Delta x = \frac{\pi}{6}$
$\displaystyle y_0 = \sin(0)$
$\displaystyle y_1 = \sin\left(\frac{\pi}{6}\right)$
$\displaystyle y_2 = \sin\left(\frac{\pi}{3}\right)$
$\displaystyle y_3 = \sin\left(\frac{\pi}{2}\right)$
$\displaystyle y_4 = \sin\left(\frac{2\pi}{3}\right)$
you have the same values as I , except you are mixing degrees with radians. don't do that.
note that you were lucky since sin(120) = sin(60) ... you listed sin(120) twice.