Exercise 7.10 · Q5: $\int_{-5}^{5}|x+2|\,dx$

Basics you need

Solution

1. Find the turning point and remove the bars

   $x + 2 \ge 0$ exactly when $x \ge -2$⁠. So on our interval

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

   At $x = -2$ both formulas give 0, so they agree there. The function $|x + 2|$ is continuous on $[-5, 5]$⁠.

   We know how to integrate $x + 2$ and $-(x + 2)$⁠, but $|x + 2|$ equals a different one of them on each side of the turning point; once the bars are removed, each side is a plain linear function. The turning point $x = -2$ is where the expression inside the bars is 0, and it lies inside $[-5, 5]$⁠, so the integral must be split there.
   Check a point on each side: at $x = -4$⁠, $x + 2 = -2 < 0$ and $|x + 2| = 2 = -(x + 2)$⁠; at $x = 3$⁠, $x + 2 = 5 > 0$ and $|x + 2| = 5$⁠.

2. Split with P2 at $c = -2$

   $$\begin{aligned}I &= \int_{-5}^{-2}|x + 2|\,dx + \int_{-2}^{5}|x + 2|\,dx \\ &= \int_{-5}^{-2}-(x + 2)\,dx + \int_{-2}^{5}(x + 2)\,dx .\end{aligned}$$
   P2 holds for any point $c$⁠; we choose $c = -2$ so that on each piece the integrand is given by one formula from Step 1. (The book proves P2 from Theorem 2: $F(c) - F(a)$ plus $F(b) - F(c)$ is $F(b) - F(a)$⁠.)
   P2, Section 7.10. Lesson: properties of definite integrals

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

   An antiderivative of $x + 2$ is $\frac{x^2}{2} + 2x$⁠. So

   $$\begin{aligned}\int_{-5}^{-2}-(x + 2)\,dx &= -\left[\frac{x^2}{2} + 2x\right]_{-5}^{-2} \\ &= -\Big[(2 - 4) - \left(\tfrac{25}{2} - 10\right)\Big] \\ &= -\left[-2 - \tfrac{5}{2}\right] = \frac{9}{2}.\end{aligned}$$
   The integrand $-(x + 2)$ is positive on $(-5, -2)$⁠, so this piece must come out positive, and it does. Keep the brackets around the value at the lower limit: subtracting $\frac{25}{2} - 10 = \frac52$ is where sign slips happen.
   $\int x\,dx = \frac{x^2}{2}$⁠, $\int 2\,dx = 2x$ (standard formula (i), Section 7.2; check: $\frac{d}{dx}\left(\frac{x^2}{2} + 2x\right) = x + 2$⁠). Theorem 2, Section 7.8.3. Lesson: the fundamental theorem

4. The right piece, on $[-2, 5]$

   $$\begin{aligned}\int_{-2}^{5}(x + 2)\,dx &= \left[\frac{x^2}{2} + 2x\right]_{-2}^{5} \\ &= \left(\tfrac{25}{2} + 10\right) - (2 - 4) \\ &= \tfrac{45}{2} + 2 = \frac{49}{2}.\end{aligned}$$
   Same antiderivative, now with the limits $-2$ and $5$⁠. Again the integrand is $\ge 0$ on the piece, and the answer is positive.
   Theorem 2, Section 7.8.3: value at the upper limit minus value at the lower limit. Lesson: the fundamental theorem

5. Add the pieces, and check with triangles

   $$I = \frac{9}{2} + \frac{49}{2} = \frac{58}{2} = 29 .$$
   Because $|x + 2| \ge 0$⁠, the integral is the area under the graph, and the region is two right-angled triangles. Their areas agree with Steps 3 and 4, which confirms the calculation.
   Area of a triangle $= \frac12 \times \text{base} \times \text{height}$⁠. Basics: area as a sum of thin strips

6. Why P7 does not help here

   The interval $[-5, 5]$ is symmetric about 0, but P7 needs $f$ to be even or odd. For $f(x) = |x + 2|$⁠:

   $$f(1) = 3, \qquad f(-1) = |{-1} + 2| = 1 .$$

   So $f(-1) \ne f(1)$ (not even) and $f(-1) \ne -f(1)$ (not odd). The graph is symmetric about $x = -2$⁠, not about the $y$-axis.

   A symmetric interval is only half of P7's condition; the function must also be symmetric about 0. One counterexample point is enough to show it is neither even nor odd. Doubling $\int_0^5$ would give $2 \cdot \frac{45}{2} = 45$⁠, which is wrong.
   P7 (Section 7.10): (i) if $f(-x) = f(x)$⁠, then $\int_{-a}^{a} f = 2\int_0^a f$⁠; (ii) if $f(-x) = -f(x)$⁠, then $\int_{-a}^{a} f = 0$⁠. Basics: symmetry and mirror images

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