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if f'(x)=3 - sqrt(4-x^2) what is f(x)?

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  • 4 years ago
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    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

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