# Calculus

Use geometry to evaluate integral[-3,3] for:
f(x)=
{ √(4-(x+1)^2), -3≤x≤1
{ |x-2| - 1, 1<x≤3

A. 2pi + 1
B. 2pi - 1
C. pi - 1
D. pi + 1

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1. y = √(4-(x+1)^2)
y^2 = 4 - (x+1)^2
that is a circle with centre (-1,0) and radius 4
so you have 1/2 of that circle ---> (1/2)π(4) = 2π
the other part is a triangle below the x-axis
with a base of 2 and a height of 1
area = (1/2)(2)(1) = 1
total area = 2π + 1

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2. typo, radius 2 I think

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3. Thanks for the catch Damon,
typo indeed since I used 2 in the calculation

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