Definite Integrals of Modulus Functions (Exercise 7.10)

This lesson evaluates definite integrals containing a modulus by splitting the interval at the point where the inside of the absolute value changes sign, using property P2 from Exercise 7.10. Three questions are worked in full, and one is also solved with a shortcut formula.

The splitting property (P2)

The core tool is the property that an integral over an interval equals the sum of the integrals over its two parts, where $c$ lies between $a$ and $b$:

$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$

For a modulus, we choose $c$ to be the point where the expression inside changes sign, and use the definition

$|u| = \begin{cases} u & u \ge 0 \\ -u & u < 0 \end{cases}$

Question 5: $\int_{-5}^{5} |x + 2|\,dx$

Here $x + 2 \ge 0$ when $x \ge -2$ and $x + 2 < 0$ when $x < -2$, so the sign changes at $c = -2$. Split there:

$\int_{-5}^{5} |x+2|\,dx = \int_{-5}^{-2} -(x+2)\,dx + \int_{-2}^{5} (x+2)\,dx$

Left piece

$\int_{-5}^{-2} -(x+2)\,dx = \left[-\left(\tfrac{x^2}{2} + 2x\right)\right]_{-5}^{-2} = -\Big[(2 - 4) - (\tfrac{25}{2} - 10)\Big] = \tfrac{9}{2}$

Right piece

$\int_{-2}^{5} (x+2)\,dx = \left[\tfrac{x^2}{2} + 2x\right]_{-2}^{5} = (\tfrac{25}{2} + 10) - (2 - 4) = \tfrac{49}{2}$

Adding the two pieces:

$\tfrac{9}{2} + \tfrac{49}{2} = \tfrac{58}{2} = 29$

Question 6: $\int_{2}^{8} |x - 5|\,dx$

Now $x - 5 \ge 0$ when $x \ge 5$ and $x - 5 < 0$ when $x < 5$, so $c = 5$, which lies between $2$ and $8$. Split there:

$\int_{2}^{8} |x-5|\,dx = \int_{2}^{5} -(x-5)\,dx + \int_{5}^{8} (x-5)\,dx$

Left piece

$\int_{2}^{5} -(x-5)\,dx = \left[-\left(\tfrac{x^2}{2} - 5x\right)\right]_{2}^{5} = -\Big[(\tfrac{25}{2} - 25) - (2 - 10)\Big] = \tfrac{9}{2}$

Right piece

$\int_{5}^{8} (x-5)\,dx = \left[\tfrac{x^2}{2} - 5x\right]_{5}^{8} = (32 - 40) - (\tfrac{25}{2} - 25) = \tfrac{9}{2}$

Adding the pieces:

$\tfrac{9}{2} + \tfrac{9}{2} = 9$

Question 5 again, using a shortcut

The same integral can be done with the formula

$\int |u|\,du = \frac{u\,|u|}{2}$

Applied to $|x+2|$:

$\int_{-5}^{5} |x+2|\,dx = \left[\frac{(x+2)\,|x+2|}{2}\right]_{-5}^{5} = \frac{7 \cdot 7}{2} - \frac{(-3)\cdot 3}{2} = \tfrac{49}{2} + \tfrac{9}{2} = 29$

which matches the answer found by splitting.

Question 18: $\int_{0}^{4} |x - 1|\,dx$

Here $x - 1 \ge 0$ when $x \ge 1$ and $x - 1 < 0$ when $x < 1$, so $c = 1$. Split there:

$\int_{0}^{4} |x-1|\,dx = \int_{0}^{1} -(x-1)\,dx + \int_{1}^{4} (x-1)\,dx$

Left piece

$\int_{0}^{1} -(x-1)\,dx = \left[-\left(\tfrac{x^2}{2} - x\right)\right]_{0}^{1} = -\big[(\tfrac{1}{2} - 1) - 0\big] = \tfrac{1}{2}$

Right piece

$\int_{1}^{4} (x-1)\,dx = \left[\tfrac{x^2}{2} - x\right]_{1}^{4} = (8 - 4) - (\tfrac{1}{2} - 1) = \tfrac{9}{2}$

Adding the pieces:

$\tfrac{1}{2} + \tfrac{9}{2} = \tfrac{10}{2} = 5$

Key takeaways

• To integrate a modulus, split the interval at the point $c$ where the inside changes sign, and use $\int_a^b = \int_a^c + \int_c^b$.
• On each piece, replace $|u|$ by $u$ where $u \ge 0$ and by $-u$ where $u < 0$, then integrate as normal.
• For a simple linear inside, the shortcut $\int |u|\,du = \tfrac{u\,|u|}{2}$ gives the same result more quickly.