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.
 A: 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.
A: To find the area enclosed you need the limits and the limits are found by finding the points where the two functions intersect; this is done by setting $f(x) =g(x)$
$f(x) = 10e^{-0.201 x}+3$
$g(x) = -x^2+12x-24$
$f(x) =g(x) \implies 10e^{-0.201x}+3 = -x^2+12x-24$
$\implies 10e^{-0.201x} = -x^2+12x-27$
This cannot be solved by algebraic manipulations, but you can find the values using numerical methods such as Newtons method
define $h(x)=10e^{-0.201x}+x^2-12x+27$
$x_1 = x_0-\frac{h(x_0)}{h'(x_0)}$
choosing $4$ as the initial guess and successive iterations 
 gets us;
$x_1 = 4 - \frac{h(4)}{h'(4)} = 4- \frac{-0.524648}{-4.899546} = 3.892$
$x_2 = 3.892 = \frac{h(3.892)}{h'(3.892)} = 3.892+0.003355 = 3.8953$
You could go for more iterations but i think this level of approximation is good enough.
To find the other root set $x_0$ equal to another guess, since you've said the root lies at $8.6940$ I'll let $x_0 =9$
$x_1 = 9-\frac{h(9)}{h'(9)}= 9-0.28883= 8.71117$
$x_2 = 8.71117-\frac{h(8.71117)}{h'(8.7117)}= 8.71117-0.017058 =8.694112$
hence you've found your other root.
So the integral is $\displaystyle\int_{3.8953}^{8.694112}(-x^2+12x-24-10e^{-0.201 x}-3)\,dx $
The above integral evaluates to $19.4914$ $\quad\bigg[$found by Wolfram alpha$\bigg]$
EDIT:
Note that I am using a calculator to do the calculations . If calculators are not allowed in your classes , then please  do not go for this method.
A: To find the intersection points, you have to solve for $x$, $f(x)=g(x)$ which, by the end, means that you are looking for the zeros of function
$$h(x)=10\, e^{-201 x/1000}+x^2-12 x+27$$ The derivative
$$h'(x)=2 x-\frac{201}{100} e^{-201 x/1000}-12$$ cancels at 
$$x_*=6+\frac{1000}{201} W\left(\frac{40401}{200000} e^{-603/500}\right)$$ where appears Lambert function. Since the argument is quite small $(\approx0.0604788)$, you can easily evaluate the result using the very first terms of the series expansion
$$W(t)=t-t^2+\frac{3 t^3}{2}-\frac{8 t^4}{3}+\frac{125 t^5}{24}+O\left(t^6\right)$$ and get $x_* \approx 6.28418$.
Now, make a Taylor series at $x=x_*$; since the first derivative is $0$ at $x=x_*$, you have 
$$h(x)=h(x_*)+\frac 12 h''(x_*)(x-x_*)^2+O\left((x-x_*)^3\right)$$ with
$$h''(x)=2+\frac{40401 }{100000}\,e^{-201 x/1000}$$ Ignoring the higher order terms and solving the quadratic gives, as approximation,
$$x_1=3.88369 \qquad \text{and} \qquad x_2=8.68468$$
Now, let us start Newton iterations with these  as starting values. The iterates would be
$$\left(
\begin{array}{cc}
 n & x_1^{(n)} \\
 0 & 3.883687134 \\
 1 & 3.895328641 \\
 2 & 3.895357513
\end{array}
\right)$$ and
$$\left(
\begin{array}{cc}
 n & x_2^{(n)} \\
 0 & 8.684680824 \\
 1 & 8.694070586 \\
 2 & 8.694052468
\end{array}
\right)$$ which are the solutions for ten significant figures.
