Question

AP Calculus Help Please!

Answer 1

4. By the fundamental theorem of calculus,

$\displaystyle\frac{\mathrm dg(x)}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}\int_{-2}^xf(t)\,\mathrm dt=f(x)$

$g(x)$ is increasing on those intervals where $g'(x)=\dfrac{\mathrm dg(x)}{\mathrm dx}>0$. So you have

$g'(x)=f(x)=\begin{cases}3&\text{for }-3\le x$

which is clearly positive for $x\in[-3,0)$, and in the second interval you have

$-x+3>0\implies x$

Together, this means $g'(x)>0$ for all $x\in[-3,3)$.

5. When $0\le x\le6$, $f(x)$ reduces to $-x+3$, so you have

$g(x)=\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt$
$g(x)=6+\left(3t-\dfrac12t^2\right)\bigg|_{t=0}^{t=x}$
$g(x)=6+3x-\dfrac12x^2$