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?
2 Answers
$\begingroup$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.
EDITOne 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.)
$\endgroup$ 3 $\begingroup$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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