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find the area of the region bounded by the line y=x^2 and curve y=1/4 (x-3)^2
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First try to find where y=y, you'll get the values where the region ends. Form an integral from where the region starts to where it ends.
y=y --> x^2=1/4*(x-3)^2
Also, make sure what curve lies on top in the region. Now you got an arsenal of hints!
i got up to that part. x=-3 and x=1
\[\int\limits_{-3}^{1} \]
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Okay, the integral for the region is going to be of (the upper curve - the lower curve)
the upper curve is x^2 and the lower curve is 1/4(x-3)^2
So take the integral of x^2 - 1/4(x-3)^2
With the bounds you calculated from y=y.
ok. i'll work on it! Thank you!
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No problem, good luck! :)
i got -8. is that right?
I'll check!
That would be correct!
thank you :)
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