The area of the region given by A=(x,y):x^2 yx+2,4-3x is:

Question

The area of the region given by $A=\{(x,y):x^2\le y\le\min\{x+2,4-3x\}\}$ is:

Accepted Answer

$x+2=4-3x\Rightarrow x=\frac12$. Region spans $x\in[-1,1]$ ($x^2=x+2$ at $x=-1$; $x^2=4-3x$ at $x=1$).

$$\int_{-1}^{1/2}(x+2-x^2)\,dx+\int_{1/2}^{1}(4-3x-x^2)\,dx=\frac94+\frac7{12}=\frac{27}{12}+\frac7{12}=\frac{17}6.$$

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