Re: HSC 2013 2U Marathon
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{\pi&space;(-1)^{2}}{4}&space;+&space;\frac{1}{2}.1.1&space;+&space;3.(-1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{\pi&space;(-1)^{2}}{4}&space;+&space;\frac{1}{2}.1.1&space;+&space;3.(-1)" title="\frac{\pi (-1)^{2}}{4} + \frac{1}{2}.1.1 + 3.(-1)" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{\pi}{4}&space;+&space;\frac{1}{2}&space;+&space;-3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{\pi}{4}&space;+&space;\frac{1}{2}&space;+&space;-3" title="\frac{\pi}{4} + \frac{1}{2} + -3" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{\pi&space;-&space;10}{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{\pi&space;-&space;10}{4}" title="= \frac{\pi - 10}{4}" /></a>
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Oh i get it! so would it beYou found the positive area under the curve. When you integrate, anything below the x-axis becomes negative
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{\pi&space;(-1)^{2}}{4}&space;+&space;\frac{1}{2}.1.1&space;+&space;3.(-1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{\pi&space;(-1)^{2}}{4}&space;+&space;\frac{1}{2}.1.1&space;+&space;3.(-1)" title="\frac{\pi (-1)^{2}}{4} + \frac{1}{2}.1.1 + 3.(-1)" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{\pi}{4}&space;+&space;\frac{1}{2}&space;+&space;-3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{\pi}{4}&space;+&space;\frac{1}{2}&space;+&space;-3" title="\frac{\pi}{4} + \frac{1}{2} + -3" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{\pi&space;-&space;10}{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{\pi&space;-&space;10}{4}" title="= \frac{\pi - 10}{4}" /></a>
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