∫ (x³cos(x/2)+(1/2))√(4-x²) dx [-2,2]
Distribute
∫ x³√(4-x²)cos(x/2) + (1/2)√(4-x²) dx [-2,2]
Now
Determine the function of
f(x) = x³√(4-x²)cos(x/2)
Is an odd or a function
f(-x) = -x³√(4-x²)cos(x/2)
Therefore the function is odd so automatically the answer for
∫ x³√(4-x²)cos(x/2) dx [-2,2] = 0
While for
(1/2)∫ √(4-x²) dx[-2,2]
f(-x) = √(4-x²)
So the function is even therefore
(1/2)•2 ∫ √(4-x²) dx [0,2]
∫ √(4-x²) dx [0,2]
x = 2sinθ
dx = 2cosθ dθ
2 ∫ √(4-4sin²θ)cosθ dθ
4 ∫ cos²θ dθ
2 ∫ 1+cos(2θ) dθ
2(θ+(1/2)sin(2θ))|[0,2]
2θ+sin(2θ)|[0,2]
2arcsin(x/2)+(1/2)x√(4-x²)|[0,2]
2arcsin(1)+0
= 2(π/2)
= π
Therefore
The answer is B.
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