Exercise 7.10 · Q6: $\int_2^8|x-5|\,dx$

Basics you need

Solution

1. Find the turning point and remove the bars

   $x - 5 \ge 0$ exactly when $x \ge 5$⁠. So on $[2, 8]$

   $$|x - 5| = \begin{cases} 5 - x, & 2 \le x \le 5,\\ x - 5, & 5 \le x \le 8.\end{cases}$$

   Both formulas give 0 at $x = 5$⁠, and $|x - 5|$ is continuous on $[2, 8]$⁠.

   On each side of the turning point the absolute value is a plain linear function we can integrate. The turning point is where the inside, $x - 5$⁠, is zero; it lies inside $[2, 8]$⁠, so the interval must be cut there.
   For $x < 5$⁠: $|x - 5| = -(x - 5) = 5 - x$⁠. Test: at $x = 3$⁠, $|3 - 5| = 2 = 5 - 3$⁠.

2. Split with P2 at $c = 5$

   $$I = \int_2^5(5 - x)\,dx + \int_5^8(x - 5)\,dx .$$
   P2 holds for any point $c$⁠; choosing the turning point makes each integrand a single formula.
   P2, Section 7.10. Lesson: properties of definite integrals

3. The left piece, on $[2, 5]$

   $$\begin{aligned}\int_2^5(5 - x)\,dx &= \left[5x - \frac{x^2}{2}\right]_2^5 \\ &= \left(25 - \tfrac{25}{2}\right) - (10 - 2) \\ &= \tfrac{25}{2} - 8 = \frac{9}{2}.\end{aligned}$$
   An antiderivative of $5 - x$ is $5x - \frac{x^2}{2}$ (its derivative is $5 - x$⁠). Theorem 2 gives upper value minus lower value.
   Standard formula (i), Section 7.2, with $n = 0$ and $n = 1$⁠; Theorem 2, Section 7.8.3. Lesson: the fundamental theorem

4. The right piece, on $[5, 8]$

   $$\begin{aligned}\int_5^8(x - 5)\,dx &= \left[\frac{x^2}{2} - 5x\right]_5^8 \\ &= (32 - 40) - \left(\tfrac{25}{2} - 25\right) \\ &= -8 + \tfrac{25}{2} = \frac{9}{2}.\end{aligned}$$
   An antiderivative of $x - 5$ is $\frac{x^2}{2} - 5x$⁠. Both values in the bracket are negative numbers, but their difference, the integral of a non-negative function, is positive.
   Theorem 2, Section 7.8.3. Lesson: the fundamental theorem

5. Add the pieces

   $$I = \frac{9}{2} + \frac{9}{2} = 9 .$$
   Since $|x - 5| \ge 0$⁠, the integral is the area under the graph: two congruent right-angled triangles, matching Steps 3 and 4.
   Area of a triangle $= \frac12 \times \text{base} \times \text{height}$⁠. Basics: area as a sum of thin strips

6. Why the two pieces are equal: the reflection behind P3

   Take $f(x) = |x - 5|$⁠, $a = 2$⁠, $b = 8$⁠, so $a + b = 10$⁠. Then

   $$\begin{aligned}f(10 - x) &= |10 - x - 5| = |5 - x| \\ &= |x - 5| = f(x) .\end{aligned}$$

   Replacing $x$ by $10 - x$ reflects the line in $x = 5$ and swaps $[2, 5]$ with $[5, 8]$⁠. Applying the same substitution $t = 10 - x$ to the right piece alone ($dt = -dx$⁠; $x = 5 \to t = 5$⁠, $x = 8 \to t = 2$⁠):

   $$\begin{aligned}\int_5^8|x - 5|\,dx &= \int_2^5|5 - t|\,dt \\ &= \int_2^5|x - 5|\,dx .\end{aligned}$$

   So $I = 2\int_2^5(5 - x)\,dx = 2 \cdot \frac92 = 9$⁠, the same answer.

   P3 itself holds for every continuous $f$⁠; its proof is the substitution $t = a + b - x$⁠, which reflects $[a, b]$ in its midpoint. Here we use that same substitution on one half only. It pays off because $f(a + b - x) = f(x)$⁠: the graph is symmetric about the midpoint $\frac{a + b}{2} = 5$⁠, so the two halves are equal and we only needed one of them. The middle step uses P1 to turn the limits round and absorb the minus sign from $dt = -dx$⁠, then P0 to rename $t$ as $x$⁠.
   $|5 - x| = |x - 5|$ because $|-u| = |u|$⁠. P0, P1 and P3, Section 7.10. Basics: symmetry and mirror images

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