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

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.
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1 Answer

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

Okay, you are looking at half of a circle.  Diameter is 4. So radius is 2. 

You are finding an area, the formula is A = pi * r^2

A = pi 2^2 = 4 pi

by Level 10 User (55.7k points)

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