Miscellaneous · Q31: $\int_1^4\big(|x-1|+|x-2|+|x-3|\big)\,dx$

Basics you need

Solution

1. The integrand is continuous on $[1, 4]$

   Each $|x - c|$ is continuous on all of $\R$ (its two formulas meet at $x = c$⁠, both giving 0). A sum of continuous functions is continuous, so $f$ is continuous on $[1, 4]$⁠, and Theorem 2 can be used on every piece.

   The absolute values create corners, not jumps. Corners do not matter for integration; only the formula changes there, which is why we split the interval.
   Chapter 5, Section 5.2: $|x|$ is continuous, and sums of continuous functions are continuous. Lesson: definite integrals

2. Remove the absolute values on each piece

   On $[1, 4]$ we always have $x \ge 1$⁠, so $|x - 1| = x - 1$ throughout. The other two bars switch at 2 and at 3.

   On $[1, 2]$⁠: $x \le 2 \le 3$⁠, so

   $$\begin{aligned} f(x) &= (x-1) + (2-x) + (3-x) \\ &= 4 - x . \end{aligned}$$

   On $[2, 3]$⁠: $x \ge 2$⁠, $x \le 3$⁠, so

   $$f(x) = (x-1) + (x-2) + (3-x) = x .$$

   On $[3, 4]$⁠: $x \ge 3$⁠, so

   $$\begin{aligned} f(x) &= (x-1) + (x-2) + (x-3) \\ &= 3x - 6 . \end{aligned}$$

   At the junctions the formulas agree: at $x = 2$⁠, $4 - 2 = 2$⁠; at $x = 3$⁠, $3 = 3\cdot 3 - 6$⁠. So it does not matter which piece the junction is given to.

   Inside each piece, the sign of every bracket $x - c$ is fixed, so each bar is replaced by the bracket itself (if it is $\ge 0$⁠) or by its negative (if it is $\le 0$⁠). Checking that neighbouring formulas agree at the junctions confirms both the algebra and the continuity of Step 1.
   $|x - c| = x - c$ for $x \ge c$⁠; $|x - c| = c - x$ for $x \le c$ (Basic 1).

3. Split the integral with P2

   Applying P2 at $c = 2$ and then at $c = 3$⁠:

   $$\begin{aligned} I &= \int_1^2 (4 - x)\,dx + \int_2^3 x\,dx \\ &\quad + \int_3^4 (3x - 6)\,dx . \end{aligned}$$
   P2 holds for any point $c$ (Section 7.10, proved from Theorem 2). Changing a function's formula at a single point does not change an integral, and here the formulas agree at the junctions anyway, so each piece is exactly the original $f$ on that interval.
   P2: $\int_a^b f = \int_a^c f + \int_c^b f$⁠. Lesson: properties of definite integrals

4. Integrate each piece

   $$\begin{aligned} \int_1^2 (4 - x)\,dx &= \Big[4x - \tfrac{x^2}{2}\Big]_1^2 \\ &= (8 - 2) - \left(4 - \tfrac12\right) = \tfrac52 . \end{aligned}$$ $$\int_2^3 x\,dx = \Big[\tfrac{x^2}{2}\Big]_2^3 = \tfrac92 - 2 = \tfrac52 .$$ $$\begin{aligned} \int_3^4 (3x - 6)\,dx &= \Big[\tfrac{3x^2}{2} - 6x\Big]_3^4 \\ &= (24 - 24) - \left(\tfrac{27}{2} - 18\right) \\ &= \tfrac92 . \end{aligned}$$

   Check by differentiating the three antiderivatives: $\frac{d}{dx}(4x - \frac{x^2}{2}) = 4 - x$⁠, $\frac{d}{dx}\frac{x^2}{2} = x$⁠, $\frac{d}{dx}(\frac{3x^2}{2} - 6x) = 3x - 6$ ✓.

   Each piece is a polynomial, so the power rule (standard formula (i), Section 7.2) and Theorem 2 finish it. In the last piece, $\frac{27}{2} - 18 = -\frac92$⁠, and subtracting a negative gives $+\frac92$⁠.
   Basics: powers of x and the power rule

5. Add, and read the answer from the graph

   $$I = \frac52 + \frac52 + \frac92 = \frac{19}{2} = 9.5 .$$
   Since $f \ge 0$⁠, the integral is the area under the graph. The graph passes through $(1, 3)$⁠, $(2, 2)$⁠, $(3, 3)$⁠, $(4, 6)$⁠, so the three trapeziums have areas $\frac{3 + 2}{2} = \frac52$⁠, $\frac{2 + 3}{2} = \frac52$ and $\frac{3 + 6}{2} = \frac92$ (each has width 1). This agrees with Step 4 without any integration.
   Basics: area as a sum of thin strips

6. Cross-check: integrate term by term

   The integral of a sum is the sum of the integrals, so we can also take each bar on its own over $[1, 4]$⁠:

   $$\begin{aligned} \int_1^4 |x - 1|\,dx &= \int_1^4 (x - 1)\,dx \\ &= \Big[\tfrac{(x-1)^2}{2}\Big]_1^4 = \tfrac92 , \end{aligned}$$ $$\begin{aligned} \int_1^4 |x - 2|\,dx &= \int_1^2 (2 - x)\,dx \\ &\quad + \int_2^4 (x - 2)\,dx \\ &= \tfrac12 + 2 = \tfrac52 , \end{aligned}$$ $$\begin{aligned} \int_1^4 |x - 3|\,dx &= \int_1^3 (3 - x)\,dx \\ &\quad + \int_3^4 (x - 3)\,dx \\ &= 2 + \tfrac12 = \tfrac52 . \end{aligned}$$

   Total: $\frac92 + \frac52 + \frac52 = \frac{19}{2}$ ✓, the same.

   Each $\int|x - c|$ is the area of one or two right triangles: for example $\int_1^4|x - 2|\,dx$ is a triangle with legs 1 and 1 (area $\frac12$⁠) plus one with legs 2 and 2 (area 2). The two methods split the same area in different ways.
   Property (III), Section 7.2.1, carried over to definite integrals. Lesson: properties of definite integrals

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