ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: ahat on July 04, 2013, 06:30:52 pm

Title: Help finding area under curve
Post by: ahat on July 04, 2013, 06:30:52 pm: y = -2 cos ( $\frac{x}{2}$ ) between the x-axis, x = $\frac{\pi}{6}$ and x = $\frac{7\pi}{6}$

I keep getting $8 - 2$ $\sqrt6$ - $2$ $\sqrt2$ $units$ ²
Title: Re: Help finding area under curve
Post by: lzxnl on July 04, 2013, 08:33:26 pm: Ok, we need to find where y > 0
cos x/2 < 0 and pi/6 < x < 7pi/6
well cos x /2 = 0 for x = pi only in that domain
and you can see that if pi/6 < x < pi, then cos x/2 > 0
so y > 0 for pi < x < 7pi/6
Your area is then $\int^{\pi}_{\pi/6} 2 \cos{\frac{x}{2}} dx - \int^\frac{7 \pi}{6}_{\pi} 2 \cos{\frac{x}{2}} dx$
An antiderivative of 2 cos x/2 is simply 4 sin x/2, so plugging all of the numbers in yields:
4 sin(pi/2) - 4 sin(pi/12) + 4 sin(pi/2) - 4 sin (7pi/12)

Now sin(pi/12) = sin(pi/3 - pi/4) = sin(pi/3)cos(pi/4) - sin(pi/4)cos(pi/3)
= $\frac{\sqrt{2}}{2} (\frac{\sqrt{3}}{2} - 1/2)$
= $\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$

and sin(7pi/12) is the same as the above except putting a plus sign where there was a minus sign
so $\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$

Putting this together, your answer should be $8 - 2\sqrt{6}$

Yes as you can see, I'm lazy with latex.
Title: Re: Help finding area under curve
Post by: ahat on July 04, 2013, 09:07:31 pm: Quote from: ahat on July 04, 2013, 06:30:52 pm
y = -2 cos ( $\frac{x}{2}$ ) between the x-axis, x = $\frac{\pi}{6}$ and x = $\frac{7\pi}{6}$

I keep getting $8 - 2$ $\sqrt6$ - $2$ $\sqrt2$ $units$ ²

Ahh damn, this is where I went wrong.
Thank you and much love ♥
Title: Re: Help finding area under curve
Post by: ahat on July 04, 2013, 09:09:07 pm: Hahahah, that quote didn't work lol.

I meant

"and sin(7pi/12) is the same as the above except putting a plus sign where there was a minus sign"

Thanks man :)

(i'm new to latex, quoting, text wraps and everything :( )