Why Does the Integral of |x^2 - 9| from 0 to 4 Require Splitting at x=3?

[Whalstib]

Homework Statement

∫ |x^2 -9| [0-4]

Homework 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...

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 ∂ ∏ ∑ ± − ÷ √ ∫ ∞ ~ ≈ ≠ ≡ ≤ ≥ °

 

[HallsofIvy]

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