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Absolute Value Integrals

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data
    ∫ |x^2 -9| [0-4]




    2. Relevant equations

    The book answer states the same EXCEPT splits into [0-3] and [3-4]. Other problems split the integral perfectly in half for absolute values...why would it differ and are there rules to figure this out? Larson's Calculus has no mention....sigh....



    3. The attempt at a solution
    I split the expression into |9x-x^3/3| [0-2] and |x^3/3 - 9x| [2-4] and get -45/3

    Book answer 64/3





    α β γ δ ε θ λ μ ν π ρ σ τ η φ χ ψ ω Γ Δ Θ Λ Π Σ Φ Ψ Ω ∂ ∏ ∑ ± − ÷ √ ∫ ∞ ~ ≈ ≠ ≡ ≤ ≥ °
     
  2. jcsd
  3. Jul 24, 2011 #2

    HallsofIvy

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    Well, it helps a lot to know what absolute value means! If [itex]x< 3[/itex], [itex]x^2< 9[/itex] so [itex]x^2- 9< 0[/itex] and [itex]|x^2- 9|= 9- x^2[/itex]. If [itex]x\ge 3[/itex], [itex]x^2\ge 9[/itex] so [itex]x^2- 9\ge 0[/itex] and [itex]|x^2- 9|= x^2- 9[/itex].
     
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