Let f : [-3,1] R be given as f(x) = casesx+6,,x^2, & -3 x0 x,,x^2, & 0 x1cases.

Question

Let $f : [-3,1] 	o \mathbb{R}$ be given as $f(x) = \begin{cases}\min\{x+6,\,x^2\}, & -3\le x\le0\ \max\{\sqrt x,\,x^2\}, & 0\le x\le1\end{cases}$. If the area bounded by $y=f(x)$ and the $x$-axis is $A$, then the value of $6A$ is equal to:

Answer: 41

Accepted Answer

On $[-3,0]$: $x+6 \le x^2$ for $x\le-2$, so $f(x)=x+6$ on $[-3,-2]$ and $f(x)=x^2$ on $[-2,0]$. On $[0,1]$: $\sqrt x\ge x^2$, so $f(x)=\sqrt x$.

$$A = \int_{-3}^{-2}(x+6)\,dx + \int_{-2}^{0}x^2\,dx + \int_0^1\sqrt x\,dx$$

$$= \left[\frac{x^2}{2}+6x\right]_{-3}^{-2} + \left[\frac{x^3}{3}\right]_{-2}^0 + \left[\frac23 x^{3/2}\right]_0^1 = \frac{41}{6}$$

$6A = 41$

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