#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 #