The area of the region bounded by the curve y | vedclass

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(A) We need to find the area bounded by $y = \max \{| x |, x | x - 2 |\}$,the $x$-axis,$x = -2$,and $x = 4$.First,let $f(x) = |x|$ and $g(x) = x|x-2|$

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$12$

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(A) We need to find the area bounded by $y = \max \{| x |, x | x - 2 |\}$,the $x$-axis,$x = -2$,and $x = 4$.First,let $f(x) = |x|$ and $g(x) = x|x-2|$.For $x \in [-2, 0]$,$f(x) = -x$ and $g(x) = x(2-x) = 2x - x^2$. Since $f(x) \ge g(x)$ in this interval,the area is $\int_{-2}^{0} (-x) dx = [-\frac{x^2}{2}]_{-2}^{0} = 0 - (-2) = 2$.For $x \in [0, 2]$,$f(x) = x$ and $g(x) = x(2-x) = 2x - x^2$. Here $f(x) \ge g(x)$,so the area is $\int_{0}^{2} x dx = [\frac{x^2}{2}]_{0}^{2} = 2$.For $x \in [2, 3]$,$f(x) = x$ and $g(x) = x(x-2) = x^2 - 2x$. Here $f(x) \ge g(x)$,so the area is $\int_{2}^{3} x dx = [\frac{x^2}{2}]_{2}^{3} = \frac{9}{2} - 2 = 2.5$.For $x \in [3, 4]$,$g(x) = x^2 - 2x$ and $f(x) = x$. Here $g(x) \ge f(x)$,so the area is $\int_{3}^{4} (x^2 - 2x) dx = [\frac{x^3}{3} - x^2]_{3}^{4} = (\frac{64}{3} - 16) - (9 - 9) = \frac{64-48}{3} = \frac{16}{3} \approx 5.33$.Wait,re-evaluating the graph: The region is bounded by $y = |x|$ for $x \in [-2, 3]$ and $y = x(x-2)$ for $x \in [3, 4]$.Area = $\int_{-2}^{0} |x| dx + \int_{0}^{3} x dx + \int_{3}^{4} (x^2 - 2x) dx = 2 + \frac{9}{2} + \frac{16}{3} = 2 + 4.5 + 5.33 = 11.83 \approx 12$.

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 . . . . . . .

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