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
Understanding the Function
We need to find the area bounded by the curve $y = \max \{|x|, x|x-2|\}$, the x-axis, from $x = -2$ to $x = 4$. First, let's define the function $f(x) = \max \{|x|, x|x-2|\}$ piecewise.
Piecewise Definition of $f(x)$
We analyze the two parts of the function:
- $|x|$ is $-x$ for $x < 0$ and $x$ for $x \ge 0$.
- $x|x-2|$ is $x(2-x) = 2x - x^2$ for $x < 2$ and $x(x-2) = x^2 - 2x$ for $x \ge 2$.
Comparing these parts in different intervals:
- For $x < 0$: $|x| = -x$ and $x|x-2| = 2x - x^2$. Since $x^2 - 3x \ge 0$ for $x < 0$, we have $|x| \ge x|x-2|$. Thus, $f(x) = -x$.
- For $0 \le x < 2$: $|x| = x$ and $x|x-2| = 2x - x^2$. We compare $x$ and $2x - x^2$. The inequality $x \ge 2x - x^2$ simplifies to $x^2 - x \ge 0$, which holds for $x \ge 1$. The inequality $2x - x^2 \ge x$ simplifies to $x^2 - x \le 0$, which holds for $0 \le x \le 1$. Thus, $f(x) = 2x - x^2$ for $0 \le x < 1$ and $f(x) = x$ for $1 \le x < 2$.
- For $x \ge 2$: $|x| = x$ and $x|x-2| = x^2 - 2x$. We compare $x$ and $x^2 - 2x$. The inequality $x \ge x^2 - 2x$ simplifies to $x^2 - 3x \le 0$, which holds for $0 \le x \le 3$. Combined with $x \ge 2$, this means $f(x) = x$ for $2 \le x \le 3$. The inequality $x^2 - 2x \ge x$ simplifies to $x^2 - 3x \ge 0$, which holds for $x \le 0$ or $x \ge 3$. Combined with $x \ge 2$, this means $f(x) = x^2 - 2x$ for $x \ge 3$.
The piecewise definition is:
$ f(x) = \begin{cases} -x & \text{if } x < 0 \\ -x^2 + 2x & \text{if } 0 \le x < 1 \\ x & \text{if } 1 \le x \le 3 \\ x^2 - 2x & \text{if } x > 3 \end{cases} $
Calculating the Area
The area is the definite integral of $f(x)$ from $x = -2$ to $x = 4$. We split the integral according to the piecewise definition and the given bounds:
Area $= \int_{-2}^{4} f(x) \, dx = \int_{-2}^{0} (-x) \, dx + \int_{0}^{1} (-x^2 + 2x) \, dx + \int_{1}^{3} x \, dx + \int_{3}^{4} (x^2 - 2x) \, dx$
Evaluating the Integrals
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Integral 1:
$ \int_{-2}^{0} (-x) \, dx = \left[ -\frac{x^2}{2} \right]_{-2}^{0} = (0) - \left( -\frac{(-2)^2}{2} \right) = 0 - (-\frac{4}{2}) = 2 $
-
Integral 2:
$ \int_{0}^{1} (-x^2 + 2x) \, dx = \left[ -\frac{x^3}{3} + x^2 \right]_{0}^{1} = \left( -\frac{1^3}{3} + 1^2 \right) - (0) = -\frac{1}{3} + 1 = \frac{2}{3} $
-
Integral 3:
$ \int_{1}^{3} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{3} = \frac{3^2}{2} - \frac{1^2}{2} = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4 $
-
Integral 4:
$ \int_{3}^{4} (x^2 - 2x) \, dx = \left[ \frac{x^3}{3} - x^2 \right]_{3}^{4} = \left( \frac{4^3}{3} - 4^2 \right) - \left( \frac{3^3}{3} - 3^2 \right) = \left( \frac{64}{3} - 16 \right) - (9 - 9) = \frac{64 - 48}{3} - 0 = \frac{16}{3} $
Total Area Calculation
Summing the results of the integrals:
Total Area $= 2 + \frac{2}{3} + 4 + \frac{16}{3} = 6 + \frac{2+16}{3} = 6 + \frac{18}{3} = 6 + 6 = 12$
The area of the region is 12.
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