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So the question is listed below as a picture, I realize I have to subtract the min and max after inputting them into the equation. So far I think it would be integral x going from 0 and going to 11/2? enter image description here

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    $\begingroup$ You have the right limits, but they are limits on $y$, not on $x$. $\endgroup$ – Brian M. Scott Dec 18 '13 at 22:51
  • $\begingroup$ You want area to be positive, so you should integrate the difference $x_{blue} - x_{red}$ (both functions of $y$) over the interval $[0, \frac{11}{2}]$. $\endgroup$ – Sammy Black Dec 18 '13 at 22:52
  • $\begingroup$ Hey, I am still not sure of how to get the answer :( is it 13.75? $\endgroup$ – Harry Johnson Dec 18 '13 at 23:47
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You can calculate the area to the right of both curves and left of the $y$-axis between $y=0$ and $y=\frac {11}2$ by integrating the given functions. Then, you can substract the results to get the area.

Also, just mirroring the image in $x=y$ or rotating it by a quarter turn may help.

EDIT One integral (the blue one) should be $\frac{121}{12}$ and the red one should be $\frac{1573}{24}$. The difference is the area. (I'm assuming we're only talking about positive areas here.)

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  • $\begingroup$ I got -13.75 + 13.75 (after inputting 11/2 for y in both equations) I am so confused $\endgroup$ – Harry Johnson Dec 18 '13 at 23:25
  • $\begingroup$ Yes, I get the difference to be -1331/24, not sure if it's correct, I am so upset with myself, I am having too much difficulty with this question! $\endgroup$ – Harry Johnson Dec 18 '13 at 23:55
  • $\begingroup$ if you omit the minus-sign, it is correct. Because you are calculating an area, it has to be positive. (For integrals in general, that isn't true, but because of the interpretation, the answer is positive here.) $\endgroup$ – Ragnar Dec 19 '13 at 0:23
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If you look sideways you can do $\int \Delta x\ dy$ to get the area. The advantage is that you always take the difference of the two curves. If you integrate $\int \Delta y \ dx$ you have regions where you need to integrate between branches of the same curve. Either way will work, but having only one type of integral simplifies things. $$\begin {align}\int_0^{11/2} 3y-y^2-(y^2-8y) dy&=\int_0^{11/2} 11y-2y^2 dy\\ &=\left.\frac {11y^2}2-\frac 23y^3\right|_0^{11/2}\\ &=\frac 13\left(\frac{11}2\right)^3\\&=\frac{1331}{24}\end {align}$$

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  • $\begingroup$ Thank you so, so, so much. I finally comprehend :) $\endgroup$ – Harry Johnson Dec 19 '13 at 0:33

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