Integration limits problem Here is a part of a problem I have a hard time with:

Let $$f(x)= 10e^{-0.201x}+3$$
Let $$g(x)= -x^2+12x-24$$
Find the area enclosed by the graphs of f and g

Here is the answer as explained by the teacher:

Finding limits $3.8953$ and $8.6940$ 
Evidence of integrating and subtracting functions 
Correct expression is....

And then he integrates both the functions with these two limits above.
He finds an area of $19.5$.
But how did he find these two limits in the first place?
Thanks.

The limits are found as the intersections of the two curves, a decaying exponential and a downward parabola. There will be no closed-form of the roots and you need to use numerical methods.
To obtain good starting estimates, you can replace the exponential (blue) by its second order development (magenta), to obtain a quadratic approximation. For convenience we will shift the origin of the coordinates to the vertex of the parabola, at $x=6$. Now with $z=x-6$,
$$-z^2+12=10e^{-0.203(z+6)}=10e^{-1.218}e^{-0.203z}+3\\
\approx10e^{-1.218}(1-0.203z+0.0206045z^2)+3.$$
Solving the quadratic equation, we find
$$x=z+6=3.879932\text{ or }x=8.68608.$$

Then you can refine with Newton.
