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

$\displaystyle y = x+3$

Intersects are -1 and 3

$\displaystyle \int_0^1 x^2-x - (x+3)$ = $\displaystyle \int_0^1 x^2-2x-3$

$\displaystyle [\frac{x^3}{3}-x^2-3x]^3_{-1}$

$\displaystyle = (9-9-9)-(-\frac{1}{3}-1+3) = -9+\frac{1}{3}+1-3$ = $\displaystyle -\frac{32}{3}$

2) y = \sqrt{x}

y = \frac{1}{2}x

x = 9

There's an intersection at x = 4, so the Area between curves is:

$\displaystyle \int^4_0\sqrt{x} - \frac{1}{2}x+$$\displaystyle \int^9_4\sqrt{x} - \frac{1}{2}x$

Shorthanding this answer: the antiderivative is $\displaystyle \frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^2}{4}$

And for 4 you get $\displaystyle \frac{16}{3}-4$, and for 9 you get $\displaystyle \frac{-9}{4}$ and for 0 it's 0, so the answer is:

$\displaystyle \frac{16}{3}-4+[\frac{-9}{4}-\frac{16}{3}-4]$

A bunch of stuff cancels and you get $\displaystyle \frac{-9}{4}$

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The two above I though I did correctly, and didn't apparently, so I don't know where I went wrong on those.

3) Find the number *b* such that the line *y* = *b* divides the region bounded by the curves *y* = *x*^2 and *y* = 4 into two regions with equal area. (Round your answer to the nearest hundredth.)

This one I'm just not sure where to get started.