The area of the region S = (x,y) : 3x^24y6x+24 is ________.

Question

The area of the region $S = \{(x,y) : 3x^2\le4y\le6x+24\}$ is \_\_\_\_\_\_\_\_.

Answer: 27

Accepted Answer

$y=\dfrac{3x^2}{4}$ and $y=\dfrac{3x}{2}+6$ meet where $3x^2=6x+24 \Rightarrow x^2-2x-8=0 \Rightarrow x=-2,4$.

$$\text{Area} = \int_{-2}^{4}\left(\frac{3x}{2}+6-\frac{3x^2}{4}\right)dx = \left[\frac{3x^2}{4}+6x-\frac{x^3}{4}\right]_{-2}^{4}$$

$= (12+24-16) - (3-12+2) = 20-(-7) = 27$

Solutions · 1