• May 25th 2011, 05:05 AM
decoy808
http://quicklatex.com/cache3/ql_4cc2...9b4c853_l3.png

get confused with the wordy ones.......do I integrate this twice, then add coordinate values ?

many thanks
• May 25th 2011, 05:08 AM
Ackbeet
You do integrate twice. In the process of so doing, you should obtain two arbitrary constants of integration. If you plug in those two points (which, incidentally, can be done while integrating, sometimes, to save effort), you will be able to determine what those arbitrary constants of integration are. Make sense? What do you get?
• May 25th 2011, 06:17 AM
decoy808
when you start to talk about the constants it gets me sidelined...but heres what i did..?

http://quicklatex.com/cache3/ql_3041...3d6b06e_l3.png
• May 25th 2011, 06:19 AM
decoy808
oh almost forogt....does it have to be in radian?
• May 25th 2011, 06:49 AM
e^(i*pi)
Your second integration is incorrect - you're also integrating the +C from the first integration. You may also have a sign error on the first

$y' = -\dfrac{1}{2}\cos(2x) +c_1$

$y = -\dfrac{1}{4}\sin(2x) + c_1x + c_2$

where $c_1 \text{ and } c_2$ are constants to be found

edit: yes, it must be in radians. Nearly all trig calc is in radians as $\dfrac{d}{dx} \sin(x) = \cos(x)$ (and all the others) work only in radians
• May 25th 2011, 08:16 AM
decoy808
must i find c1 at y'
or find c1 and c2 at y
• May 25th 2011, 08:29 AM
e^(i*pi)
The question says that "the function passes through" your two points - this would be y rather than y'.