Answer
a. $F(x)=\displaystyle \frac{2}{3}x^{3/2}-\frac{16}{3}$
b. $f(x)=\sqrt{x}$
Work Step by Step
a.
$F(x)=\displaystyle \int_{4}^{x}\sqrt{t}dt=\int_{8}^{x}t^{1/2}dt$
$=[\displaystyle \frac{2}{3}t^{3/2}]_{4}^{x}$
$=\displaystyle \frac{2}{3}[x^{3/2}-(2^{2})^{3/2}]$
$=\displaystyle \frac{2}{3}[x^{3/2}-8]$
$=\displaystyle \frac{2}{3}x^{3/2}-\frac{16}{3}$
b.
$\displaystyle \frac{d}{dx}[\frac{2}{3}x^{3/2}-\frac{16}{3}]=x^{1/2}-0=\sqrt{x}$
