The area of the region bounded by the curve $y=\max\{|x|, x|x-2|\}$, the $x$-axis and the lines $x=-2$ and $x=4$ is equal to
Solution:
$$y = \max\{|x|, x|x-2|\}$$
$$\text{Let } f_1(x) = |x| \text{ and } f_2(x) = x|x-2|$$
$$\text{For } x \in [-2, 0]: \quad f_1(x) = -x, \quad f_2(x) = 2x-x^2 \quad \Rightarrow \max \text{ is } -x$$
$$\text{For } x \in [0, 2]: \quad f_1(x) = x, \quad f_2(x) = 2x-x^2$$
$$x = 2x – x^2 \Rightarrow x^2 – x = 0 \Rightarrow x(x-1) = 0 \Rightarrow x=0, 1$$
$$\text{In } (0, 1): 2x-x^2 > x \quad \Rightarrow \max \text{ is } 2x-x^2$$
$$\text{In } (1, 2): x > 2x-x^2 \quad \Rightarrow \max \text{ is } x$$
$$\text{For } x \in [2, 4]: \quad f_1(x) = x, \quad f_2(x) = x^2-2x$$
$$x = x^2 – 2x \Rightarrow x^2 – 3x = 0 \Rightarrow x(x-3) = 0 \Rightarrow x=0, 3$$
$$\text{In } (2, 3): x > x^2-2x \quad \Rightarrow \max \text{ is } x$$
$$\text{In } (3, 4): x^2-2x > x \quad \Rightarrow \max \text{ is } x^2-2x$$
$$\text{Required Area } A = \int_{-2}^{4} y \, dx$$
$$A = \int_{-2}^{0} (-x) \, dx + \int_{0}^{1} (2x – x^2) \, dx + \int_{1}^{3} x \, dx + \int_{3}^{4} (x^2 – 2x) \, dx$$
$$A = \left[ -\frac{x^2}{2} \right]_{-2}^{0} + \left[ x^2 – \frac{x^3}{3} \right]_{0}^{1} + \left[ \frac{x^2}{2} \right]_{1}^{3} + \left[ \frac{x^3}{3} – x^2 \right]_{3}^{4}$$
$$A = (0 – (-2)) + \left(1 – \frac{1}{3}\right) + \left(\frac{9}{2} – \frac{1}{2}\right) + \left(\left(\frac{64}{3} – 16\right) – \left(\frac{27}{3} – 9\right)\right)$$
$$A = 2 + \frac{2}{3} + 4 + \left(\frac{16}{3} – 0\right)$$
$$A = 6 + \frac{18}{3}$$
$$A = 6 + 6$$
$$A = 12$$
Ans. (12)