#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
Explanation:
Let us define the functions
# 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) < 0 => F(x) # is decreasing
So let us find
# 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 => 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
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) = 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 #