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definite integral of sqrt(16-x^2) between -4 and 0

0 votes

I know the answer is 4*pi but I have no idea where the pi came from or even what to do.  I tried the substitution method but it got really messy and netted me nothing.

asked Nov 30, 2013 in CALCULUS by linda Scholar

1 Answer

0 votes

Given Sqrt(16-x^2)

Given limits are a = -4,b = 0

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answered Jan 29, 2014 by david Expert

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