#### Answer

$12\ln(6) - 10 \approx 11.501$

#### Work Step by Step

We look for the intersection point between the two equations to find the upper bound on the integral.
$12 - e^x = e^x$
$x = \ln(6)$
We also know $12-e^x$ is the upper curve so the integral will be
$\int_0^{\ln(6)} 12 - e^x - e^x dx$
$= [12x - 2e^x]_0^{\ln(6)}$
$= 12\ln(6) - 12 + 2 = 12\ln(6) - 10 \approx 11.501$