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Find the area under the graph of f over [-2,2] f(x)={4-x^2, if x<0 4, if x>0
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Sorry, \[4 if x \ge 0\]
I got 9.3, not sure where I went wrong.
It looks as if you will need to set up 2 separate integrals. First, integrate the top function (for x<0) on [-2,0]. --> 16/3 Then, add it to the integral of the bottom part on [0,2]. --> 8 Your exact answer is 40/3, or 13.333333333333...
YES. 40/3
Okay :O So then for the area bounded by y=x^2 and y=4
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Intersects are (-2,4) and (2,4) so then my interval is -[2,2]
\[\int\limits_{-2}^{2}x^2dx-\int\limits_{-2}^{2}4dx\]
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