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if f'(x)=3 - sqrt(4-x^2) what is f(x)?
1 Answer
- germanoLv 74 years agoFavorite Answer
Hello,
we need to evaluate the indefinite integral:
f(x) = ∫ f '(x) dx =
∫ [3 - √(4 - x²)] dx =
(splitting into two integrals and factoring the constant out)
3 ∫ dx - ∫ √(4 - x²) dx =
3x - ∫ √(4 - x²) dx (##)
let's rewrite the remaining integral as:
∫ √(2² - x²) dx =
let:
x = 2sinθ
dx = 2cosθ dθ
yielding, by trig substitution:
∫ √(2² - x²) dx = ∫ {√[2² - (2sinθ)²]} 2cosθ dθ =
∫ [√(2² - 2²sin²θ)] 2cosθ dθ =
∫ {√[2²(1 - sin²θ)]} 2cosθ dθ =
(applying the identity 1 - sin²θ = cos²θ)
∫ [√(2²cos²θ)] 2cosθ dθ =
∫ (2cosθ) 2cosθ dθ =
∫ 4cos²θ dθ =
(considering the argument θ as (2θ)/2)
∫ 4cos²[(2θ)/2] dθ =
let's apply the half-angle identity cos²(α/2) = (1 + cos α)/2:
∫ 4 {[1 + cos(2θ)] /2} dθ =
∫ 2 [1 + cos(2θ)] dθ =
∫ [2 + 2cos(2θ)] dθ =
(splitting into two integrals)
∫ 2 dθ + ∫ 2cos(2θ) dθ =
2θ + ∫ cos(2θ) 2 dθ =
(being 2, in the remaining integral, the derivative of the argument 2θ)
2θ + ∫ cos(2θ) d(2θ) =
(applying the integration rule ∫ cos[f(x)] d[f(x)] = sin[f(x)] + C)
2θ + sin(2θ) + C =
(applying the double-angle identity sin(2θ) = 2sinθ cosθ)
2θ + 2sinθ cosθ + C
let's now recall that x = 2sinθ; hence:
sinθ = x/2
θ = arcsin(x/2)
cosθ = √(1 - sin²θ) = √[1 - (x/2)²] = √[1 - (x²/4)] = √[(4 - x²)/4] = [√(4 - x²)] /2
then, substituting back:
2θ + 2sinθ cosθ + C = 2arcsin(x/2) + 2(x/2) {[√(4 - x²)] /2} + C =
2arcsin(x/2) + (x/2) √(4 - x²) + C =
2arcsin(x/2) + (1/2)x √(4 - x²) + C
finally let's insert this result into the above (##) expression, yielding:
3x - [2arcsin(x/2) + (1/2)x √(4 - x²)] + C =
3x - 2arcsin(x/2) - (1/2)x √(4 - x²)] + C
in conclusion:
f(x) = 3x - 2arcsin(x/2) - (1/2)x √(4 - x²)] + C
I hope it's helpful