int_0^x (t^2 -6t+8) dt where x belongs to all real number [0,infinity). Find the intervals where the function is decreasing?

1 Answer
Feb 23, 2017

2 < x < 4

Explanation:

Let us define the functions f(x) and by F(x).

f(x) \= x^2 - 6x+8

F(x) = int_0^x \ (t^2-6t+8) \ dt \ \ \ \ where x in [0,oo)
\ \ \ \ \ \ \ \= int_0^x \ f(t) \ dt

Then by the definition of an decreasing function we can state that if the derivative F'(x) satisfies F'(x)<0 on a closed interval then F(x) is decreasing on that interval. So if

F'(x) < 0 => F(x) is decreasing

So let us find F'(x)

F'(x) = d/dx F(x)
\ \ \ \ \ \ \ \ \ = d/dx int_0^x \ (t^2-6t+8) \ dt
\ \ \ \ \ \ \ \ \ = d/dx int_0^x \ f(t) \ dt
\ \ \ \ \ \ \ \ \ = f(x)

By the Fundamental Theorem of Calculus.

And so our condition F'(x) < 0 is satisfied if:

f(x) < 0 => x^2 - 6x+8 < 0
:. (x-4)(x-2) < 0
:. 2 < x < 4

graph{x^2-6x+8 [-3.625, 10.425, -2.19, 4.834]}

And hence we can conclude that the function F(x) is decreasing if the condition 2 < x < 4 is satisfied

Interpretation/Analysis
In order to interpret the above result let use evaluate the integral and from an explicit expression for the function;

F(x) = int_0^x \ (t^2-6t+8) \ dt
\ \ \ \ \ \ \ = [1/3t^3-3t^2+8t]_0^x \
\ \ \ \ \ \ \ = 1/3x^3-3x^2+8x

Here is the graph of the curve:
graph{1/3x^3-3x^2+8x [-3.88, 8.61, 2.554, 8.797]}

And it should be clear (using the definition of a decreasing function) that the function is decreasing at all point where F'(x) < 0, and F'(x)=0 corresponds to the critical points. We find F'(x) by differentiating:

F'(x) = x^2-6x+8

which is the integrand (which we found above using the FTOC) , and to find the critical points (max/min) we require:

F'(x) = 0 => x=2,4

and in order for the function to be decreasing we need:

F'(x) < 0 => 2 < x < 4