Question

`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?

Answer 1

`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`