Calculate numerically the area

[mathmari]

Hey!

The functions \begin{equation*}f(x)=\frac{1}{3}(x-2)^2e^{x/3} \ \text{ and } \ p(x)=-\frac{2}{3}x+\frac{4}{3}\end{equation*} have exactly two real intersection points at the region $x\geq 0$.

Calculate numerically the area that is between the graphs of these two functions, with accuracy of two decimal digits. For that do we have to claulate numerically the intersections points, using for example Newton's methos? (Wondering)

 

[I like Serena]

Hey mathmari!

We could, but perhaps the intersection points are "nice" points.

What are the potentially nice points? (Wondering)

 

[mathmari]

We could, but perhaps the intersection points are "nice" points.

What are the potentially nice points? (Wondering)

You mean checking it from the graph? Then we have the intersection points $x=0$ and $x=2$, right? (Wondering)

 

[I like Serena]

You mean checking it from the graph? Then we have the intersection points $x=0$ and $x=2$, right?

That works. (Nod)

Note that $f(x)=\frac{1}{3}(x-2)^2e^{x/3}$ is hard to match with a line, since it has an exponential function in it.
However, there are 2 special cases.
1. When $(x-2)=0$ the function is zero because the polynomial factor is zero.
2. When $x=0$ the exponential function is $1$.

Turns out that those specials cases coincide with the points of intersection. (Thinking)

 

[mathmari]

That works. (Nod)

Note that $f(x)=\frac{1}{3}(x-2)^2e^{x/3}$ is hard to match with a line, since it has an exponential function in it.
However, there are 2 special cases.
1. When $(x-2)=0$ the function is zero because the polynomial factor is zero.
2. When $x=0$ the exponential function is $1$.

Turns out that those specials cases coincide with the points of intersection. (Thinking)

So this is an other approach to give the intersection points, isn't? (Wondering)

To calculate numerically the area is referring to a numerical integration method and not a numerical method to find the roots, right? (Wondering)

 

[I like Serena]

So this is an other approach to give the intersection points, isn't?

To calculate numerically the area is referring to a numerical integration method and not a numerical method to find the roots, right?

Yep. (Nod)

 

[HallsofIvy]

I think Klaas Vanaarsen's point was that there are not many values that will make an exponential and a polynomial the same. But one would be to make the coefficient of the exponential 0 so the exponential disappears. The coefficient is x- 2 so we try x= 2 and find that it works. The other is the value x= 0 because then $$e^{x/2}= e^0= 1$$, an easy number!

 

[mathmari]

Ok! I got it so far! (Yes)

Which is the best numerical integration method for this case? (Wondering)

 

[I like Serena]

Which is the best numerical integration method for this case?

Looks like a case for the trapezoidal rule. (Thinking)