Let for x in R; f(x) = frac{x+|x|}{2} and g(x | vedclass

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(A) Given $f(x) = \frac{x+|x|}{2} = \begin{cases} x, & x \geq 0 \ 0, & x < 0 \end{cases}$.Given $g(x) = \begin{cases} x^2, & x \geq 0 \ x, & x < 0 \

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$72$

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(A) Given $f(x) = \frac{x+|x|}{2} = \begin{cases} x, & x \geq 0 \ 0, & x Given $g(x) = \begin{cases} x^2, & x \geq 0 \ x, & x Then $(f \circ g)(x) = f(g(x)) = \begin{cases} g(x), & g(x) \geq 0 \ 0, & g(x) For $x \geq 0$,$g(x) = x^2 \geq 0$,so $f(g(x)) = x^2$.For $x Thus,$y = (f \circ g)(x) = \begin{cases} x^2, & x \geq 0 \ 0, & x The line is $2y - x = 15$,or $y = \frac{x+15}{2}$.Intersection of $y = x^2$ and $y = \frac{x+15}{2}$ for $x \geq 0$:$x^2 = \frac{x+15}{2} \implies 2x^2 - x - 15 = 0 \implies (2x+5)(x-3) = 0$. Since $x \geq 0$,$x = 3$.The area is bounded by $y = 0$,$y = \frac{x+15}{2}$,and $y = x^2$.Area = $\int_{-15}^{0} \frac{x+15}{2} dx + \int_{0}^{3} (\frac{x+15}{2} - x^2) dx$.Area = $[\frac{x^2}{4} + \frac{15x}{2}]_{-15}^{0} + [\frac{x^2}{4} + \frac{15x}{2} - \frac{x^3}{3}]_{0}^{3}$.Area = $(0 - (\frac{225}{4} - \frac{225}{2})) + ((\frac{9}{4} + \frac{45}{2} - 9) - 0)$.Area = $\frac{225}{4} + \frac{9+90-36}{4} = \frac{225+63}{4} = \frac{288}{4} = 72$.

Let for $x \in R$; $f(x) = \frac{x+|x|}{2}$ and $g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \geq 0 \end{cases}$. Then the area bounded by the curve $y = (f \circ g)(x)$ and the lines $y = 0$,$2y - x = 15$ is equal to $...........$.

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