Integration limits problem

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.