The area (in sq.

Question

The area (in sq. units) of the region $A = \{(x,y)\in\mathbb R	imes\mathbb R \mid 0\le x\le3,\ 0\le y\le4,\ y\le x^2+3x\}$ is:

Accepted Answer

$x^2+3x=4 \Rightarrow x=1$ (taking the positive root in range).

$$\text{Area} = \int_0^1(x^2+3x)\,dx + \int_1^3 4\,dx = \left[\frac{x^3}{3}+\frac{3x^2}{2}\right]_0^1 + [4x]_1^3$$

$$= \left(\frac13+\frac32\right) + 8 = \frac{59}{6}$$

Solutions · 1