Answer:
[tex]\large \boxed{\sf \ \ 35.805 \ \ }[/tex]
Step-by-step explanation:
Hello,
First of all we need to find the intersection points of y = 12 and
[tex]y=f(x)=e^x+e^{-x}[/tex]
We need to solve the following equation.
[tex]e^x+e^{-x}=12\\\\\text{*** We multiply by }e^x\text{ both side ***}\\\\\left(e^x\right)^2+1=12e^x\\\\\text{*** Let's note } X =e^x\text{, it comes *** }\\\\X^2-12X+1=0[/tex]
[tex]\Delta=b^2-4ac=12^2-4=140=2^2\cdot 35\\\\X_1=\dfrac{12-2\sqrt{35}}{2}=6-\sqrt{35}\\\\X_2=\dfrac{12+2\sqrt{35}}{2}=6+\sqrt{35}\\[/tex]
And, we take the greater solution to solve:
[tex]X=e^x=6+\sqrt{35}<=>\boxed{x=ln(6+\sqrt{5})}[/tex]
Let's note it a.
Let's compute the integral.
We need to compute the following (because 12-f(x) is pair the integral that we are looking for is)
[tex]\displaystyle 2\int\limits^a_{0} {\left(12-(e^x+e^{-x})\right)} \, dx =12*(2a)-2\int\limits^a_{0} {(e^x+e^{-x})} \, dx \\\\=24a-2[e^x-e^{-x}]_{0}^a=24a-2(e^a-e^{-a})[/tex]
We can replace a by the value we already found.
[tex]24\cdot ln(6+\sqrt{35})-2\left( 6+\sqrt{35}-\dfrac{1}{6+\sqrt{35}}\right)\\\\=24\cdot ln(6+\sqrt{35})-2\left(6+\sqrt{35}-\dfrac{\sqrt{35}-6}{(\sqrt{35}-6)(\sqrt{35}+6)}\right)\\\\=24\cdot ln(6+\sqrt{35})-2\left(\dfrac{29*6+29*\sqrt{35}-\sqrt{35}+6}{29}\right)\\\\\\=24\cdot ln(6+\sqrt{35})-2\left(\dfrac{180+28\sqrt{35}}{29}\right)\\\\[/tex]
= 59.46933...-26.66432...=35.8050...
So the answer is [tex]\boxed{\sf \ \ 35.805 \ \ }[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
