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**Open that Hampster!** 1) $\displaystyle y = |x|$ and $\displaystyle y = x^2-6$

Using the definition of absolute value of x as a piecewise function, the intersects are -3 and 3.

$\displaystyle \int^3_{-3} (|x| - x^2-6)dx = \int^0_{-3}(-x-x^2-6)dx + \int^3_0 (x-x^2-6)dx$

$\displaystyle \frac{-x^2}{2}-\frac{x^3}{3}-6x |^0_{-3} + \frac{x^2}{2} -\frac{x^3}{3}-6x|^3_0$

$\displaystyle 0 - (\frac{9}{2} - \frac{27}{3}-18)+(\frac{9}{2}-\frac{27}{3}-18)-0$

$\displaystyle 0 -(\frac{-45}{2})+(\frac{-45}{2})$

But that can't be right....

2) $\displaystyle y = 7x^2$ and $\displaystyle y = \frac{1}{7}x$

This one I think i could do if I could get started. But I keep end up having to find the square root of 1/7, which I can't seem to figure out how to express as a fraction. Maybe just a may problem, but I don't know.