Let y₁=eˣand y₂=12-eˣ.
First we find out where they intersect, y₁=y₂:
eˣ=12-eˣ, 2eˣ=12, eˣ=6, x=ln(6).
y₁=y₂=6.
As x→−∞, y₁→0. The x-intercept of y₂ is eˣ=12, x=ln(12).
The line x=ln(6) divides the area into two parts. On the right side of x=ln(6), y₂ is the only curve contributing to the area, while on the left side y₁ alone contributes to the area:
PART 1=∫[ln(6),ln(12)](12-eˣ)dx=(12x-eˣ)[ln(6),ln(12)]=
12ln(12)-12-(12ln(6)-6)=12ln(12/6)-6=12ln(2)-6.
PART 2=∫[-∞,ln(6)]eˣdx=(eˣ)[-∞,ln(6)]=6.
So the total area is the sum of the parts:=12ln(2)=8.3178 approx.