Let, for x R, f(x)=x+|x|2 and g(x)=casesx, & x

Question

Let, for $x\in\mathbb R$, $f(x)=\dfrac{x+|x|}{2}$ and $g(x)=\begin{cases}x, & x<0\x^2, & x\ge0\end{cases}$. Then the area bounded by the curve $y=(f\circ g)(x)$ and the lines $y=0,\ 2y-x=15$ is equal to \_\_\_\_\_\_\_\_.

Answer: 72

Accepted Answer

$f(x)=x$ for $x\ge0$, $0$ for $x<0$. Since $g(x)<0$ for $x<0$ and $g(x)\ge0$ for $x\ge0$:

$$(f\circ g)(x)=\begin{cases}x^2,&x\ge0\0,&x<0\end{cases}.$$

Line: $y=\dfrac{x+15}{2}$, meeting $y=0$ at $x=-15$ and meeting $y=x^2$ at $x=3$ ($2x^2-x-15=0\Rightarrow x=3$ or $-2.5$; only $x=3$ is relevant since $(f\circ g)=x^2$ only for $x\ge0$).

$$\text{Area}=\int_{-15}^0\frac{x+15}{2}\,dx+\int_0^3\left[\frac{x+15}{2}-x^2\right]dx=\frac{225}{4}+\frac{63}{4}=72.$$

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