The area (in sq.

Question

The area (in sq. units) of the region $\left\{(x,y) : 0\le y\le x^2+1,\ 0\le y\le x+1,\ \dfrac12\le x\le2\right\}$ is:

Accepted Answer

$x^2+1=x+1 \Rightarrow x=0,1$; on $\left[\dfrac12,1\right]$, $x^2+1$ is the binding (smaller) upper bound, and on $[1,2]$, $x+1$ is binding.

$$A = \int_{1/2}^{1}(x^2+1)\,dx + \int_1^2(x+1)\,dx = \left[\frac{x^3}{3}+x\right]_{1/2}^1 + \left[\frac{x^2}{2}+x\right]_1^2$$

$$= \left(\frac43-\frac{13}{24}\right) + \frac52 = \frac{79}{24}$$

Solutions · 1