Example 28: $\int_{-1}^{2}\left|x^3-x\right|dx$

Basics you need

Solution

1. Factor and find the signs of $x^3 - x$

   $$x^3 - x = x\left(x^2 - 1\right) = x(x - 1)(x + 1).$$

   It is 0 at $x = -1, 0, 1$⁠. Between these zeros the sign is fixed, and we read it from the factors:

   Interval	$x$	$x - 1$	$x + 1$	$x^3 - x$
   $[-1, 0]$	$\le 0$	$< 0$	$\ge 0$	$\ge 0$
   $[0, 1]$	$\ge 0$	$\le 0$	$> 0$	$\le 0$
   $[1, 2]$	$> 0$	$\ge 0$	$> 0$	$\ge 0$

   For example at $x = -\frac12$⁠: $-\frac18 + \frac12 = \frac38 > 0$⁠; at $x = \frac12$⁠: $\frac18 - \frac12 = -\frac38 < 0$⁠; at $x = \frac32$⁠: $\frac{27}{8} - \frac32 = \frac{15}{8} > 0$⁠.

   The absolute value only changes its formula where $x^3 - x$ changes sign, so we must know exactly where that happens. Factoring shows the zeros; the table shows the sign in between. These are the book's three statements "$x^3 - x \ge 0$ on $[-1, 0]$", "$\le 0$ on $[0, 1]$", "$\ge 0$ on $[1, 2]$".
   On $[-1, 0]$ there are two non-positive factors ($x$ and $x - 1$⁠), so the product is $\ge 0$⁠.

2. Split the integral by P2 and remove the bars

   The integrand $|x^3 - x|$ is continuous on $[-1, 2]$ (a composition of continuous functions). Cutting at 0 and at 1 (P2 used twice):

   $$\begin{aligned}\int_{-1}^{2}|x^3 - x|\,dx = {}&\int_{-1}^{0}(x^3 - x)\,dx \\ &+ \int_0^1 -(x^3 - x)\,dx \\ &+ \int_1^2 (x^3 - x)\,dx.\end{aligned}$$

   The middle piece is $\int_0^1 (x - x^3)\,dx$⁠.

   On each piece the sign is known (Step 1), so $|x^3 - x|$ has a single formula there: itself where it is $\ge 0$⁠, its negative where it is $\le 0$⁠. At the cut points the two formulas agree (both give 0), so it does not matter which piece a cut point belongs to.
   Property P2, Section 7.10, with $c = 0$ and then $c = 1$⁠. Lesson: properties of definite integrals

3. One antiderivative serves all three pieces

   Let $G(x) = \frac{x^4}{4} - \frac{x^2}{2}$⁠, so $G'(x) = x^3 - x$⁠. Its values at the cut points:

   $$\begin{aligned}G(-1) &= \tfrac14 - \tfrac12 = -\tfrac14, & G(0) &= 0, \\ G(1) &= \tfrac14 - \tfrac12 = -\tfrac14, & G(2) &= 4 - 2 = 2.\end{aligned}$$
   Every piece is $\pm(x^3 - x)$⁠, so the same $G$ (with a minus sign for the middle piece) does all the work. Computing the four values once avoids repeated arithmetic.
   Power rule: $\int x^3dx = \frac{x^4}{4}$⁠, $\int x\,dx = \frac{x^2}{2}$⁠. Basics: the power rule

4. Evaluate each piece

   $$\begin{aligned}\int_{-1}^{0}(x^3 - x)\,dx &= G(0) - G(-1) = \tfrac14, \\ \int_0^1 (x - x^3)\,dx &= -\left[G(1) - G(0)\right] = \tfrac14, \\ \int_1^2 (x^3 - x)\,dx &= G(2) - G(1) = 2 + \tfrac14 = \tfrac94.\end{aligned}$$
   Each piece is positive, as it must be: the integrand $|x^3 - x|$ is never negative. A negative piece here would signal a sign error in removing the bars.
   Theorem 2, Section 7.8.3: F(upper) minus F(lower). Lesson: the fundamental theorem

5. Add the pieces

   $$\int_{-1}^{2}|x^3 - x|\,dx = \frac14 + \frac14 + \frac94 = \frac{11}{4} = 2.75.$$

   The book arrives at the same total by expanding all the brackets:

   $$\begin{aligned}&-\tfrac14 + \tfrac12 + \tfrac12 - \tfrac14 + 2 - \tfrac14 + \tfrac12 \\ &= \tfrac32 - \tfrac34 + 2 = \tfrac{11}{4}.\end{aligned}$$
   The figure shows why the absolute value matters: without it, the gold dip would count as $-\frac14$ and cancel the first hump, giving only $\frac94$⁠.
   Area is always counted positive; a signed integral counts parts below the axis as negative. Basics: area as a sum of thin strips

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