Answer
$\displaystyle \frac{3}{2}\ln 2$
Work Step by Step
Area between curves $f(x)$ and $g(x)$, where $g(x)\geq f(x)$ on $[a,b], $ is
$A=\displaystyle \int_{a}^{b}[g(x)-f(x)]dx$
On $[-\pi/4,0]$ the graph of tan is below the x-axis
On $[0,\pi/3]$, it is above the x-axis.
$A=\displaystyle \int_{-\pi/4}^{0}(-\tan x)dx+\int_{0}^{\pi/3}\tan xdx$
See summary before example 4,$ \displaystyle \quad \int\tan udu=\ln|\sec u|+C$
$A=-[\ln|\sec x|]_{-\pi/4}^{0}-[\ln|\sec x|]_{0}^{\pi/3}$
$=-\ln 1+\ln\sqrt{2}-\ln 2+\ln 0$
$=\displaystyle \frac{1}{2}\ln 2+\ln 2$
$=\displaystyle \frac{3}{2}\ln 2$
