The area of the region given by A=(x,y):x2≤y≤min{x+2,4−3x} is:

A
$\frac{31}{8}$
B
$\frac{17}{6}$
C
$\frac{19}{6}$
D
$\frac{27}{8}$

Solution:

$x^{2} = x + 2 \Rightarrow x = - 1 4 - 3 x = x + 2 \Rightarrow x = \frac{1}{2} 4 - 3 x = x^{2} \Rightarrow x = 1$

$A = \int_{- 1}^{\frac{1}{2}} (x + 2 - x^{2}) d x + \int_{\frac{1}{2}}^{1} (4 - 3 x - x^{2}) d x A = {(\frac{x^{2}}{2} + 2 x - \frac{x^{3}}{3})}_{- 1}^{\frac{1}{2}} + {(4 x - \frac{3 x^{2}}{2} - \frac{x^{3}}{3})}_{\frac{1}{2}}^{1} = \frac{27}{12} + \frac{7}{12} = \frac{34}{12} = \frac{17}{6}$

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

Solution:

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