Chapter 6 - Applications of the Integral - 6.1 Area Between Two Curves - Exercises - Page 286: 10

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

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 $\int_0^{\ln(6)} 12 - e^x - e^x dx$ $= [12x - 2e^x]_0^{\ln(6)}$ $= 12\ln(6) - 12 + 2 = 12\ln(6) - 10 \approx 11.501$