Attachment 20821

I've only learned the area between two curves and integration along the y-axis...I'm really confused on how to start this one. Do I find the slope of the lines and go from there? Any help would be greatly appreciated

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- Feb 15th 2011, 02:43 PMbcahmelFind area (image included)
Attachment 20821

I've only learned the area between two curves and integration along the y-axis...I'm really confused on how to start this one. Do I find the slope of the lines and go from there? Any help would be greatly appreciated - Feb 15th 2011, 02:47 PMe^(i*pi)
I would take the whole area and subtract the non shaded parts

$\displaystyle \displaystyle \int^{\pi /2}_0 \cos(x) \,dx - \left( \int^{\pi /6}_0 \cos(x) \,dx + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right)$

in case you're wondering the last term is the area of a triangle

Don't quote me on it though, someone may have a better answer! - Feb 15th 2011, 03:31 PMArchie Meade
You could also calculate

$\displaystyle \displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}c osxdx+\frac{1}{2}\frac{\pi}{6}\frac{\sqrt{3}}{2}-\frac{1}{2}\frac{1}{2}\frac{\pi}{3}$

In other words, the area under the curve from $\displaystyle \frac{\pi}{6}\rightarrow\frac{\pi}{3}$

then add in the area of the upper triangle and subtract the lower one..

@ e^(i*pi)

the last part isn't a triangle (the curve is cosx there).