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Mathematics 19 Online
OpenStudy (anonymous):

find the area of the region bounded by the line y=x^2 and curve y=1/4 (x-3)^2

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

y=y --> x^2=1/4*(x-3)^2

OpenStudy (anonymous):

Also, make sure what curve lies on top in the region. Now you got an arsenal of hints!

OpenStudy (anonymous):

i got up to that part. x=-3 and x=1

OpenStudy (anonymous):

\[\int\limits_{-3}^{1} \]

OpenStudy (anonymous):

Okay, the integral for the region is going to be of (the upper curve - the lower curve)

OpenStudy (anonymous):

the upper curve is x^2 and the lower curve is 1/4(x-3)^2

OpenStudy (anonymous):

So take the integral of x^2 - 1/4(x-3)^2

OpenStudy (anonymous):

With the bounds you calculated from y=y.

OpenStudy (anonymous):

ok. i'll work on it! Thank you!

OpenStudy (anonymous):

No problem, good luck! :)

OpenStudy (anonymous):

i got -8. is that right?

OpenStudy (anonymous):

I'll check!

OpenStudy (anonymous):

That would be correct!

OpenStudy (anonymous):

thank you :)

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