Area Under Mod!

To get a gist of how the graph will look like, we write the equation as

$|y| = \left| \dfrac{x}{4} \right| - |x-60|$

Since LHS $\ge$ 0, so similar goes for the RHS i.e.,

$\left| \dfrac{x}{4} \right| \ge |x-60|$

---

• Case 1 : $x < 0$

This gives $- \dfrac{x}{4} \ge 60 - x \implies x \ge 80$ which is impossible in this case.

• Case 2 : $0 \le x < 60$

This gives $\dfrac{x}{4} \ge 60 - x \implies x \ge 48$ so we get the interval $x \in \mathbb{N} : 48 \le x < 60$ .

• Case 3 : $x \ge 60$

This gives $\dfrac{x}{4} \ge x - 60 \implies x \le 80$ so we get the interval $x \in \mathbb{N} : 60 \le x \le 80$ .

---

Combining all the above results we find that the equation is valid for $x \in \mathbb{N} : 48 \le x \le 80$ . Also within the interval, the function has a turning point at $x=60$ .

• For $48 \le x \le 60$ : $|y| = \dfrac{5x}{4} - 60$

• For $60 \le x \le 80$ : $|y| = 60 - \dfrac{3x}{4}$

Since $y$ can be both positive and negative (including 0 as well) the graph forms symmetrically above and below the x-axis. The required area is therefore, double of the area above the axis.

Thus

$A = 2 \left[ \int_{48}^{60} \left( \dfrac{5x}{4} - 60 \right) \,dx + \int_{60}^{80} \left( 60 - \dfrac{3x}{4} \right) \,dx \right] = \boxed{480}$