#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?

Question

Emilia Aguilar · Accepted Answer

#2 < x < 4#

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 Calculus Fundamental Theorem.

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

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

Interpretation/Analysis Let's evaluate the integral and start with an explicit expression for the function to understand the above result.

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

Graph{1/3x^3-3x^2+8x [-3.88, 8.61, 2.554, 8.797]} is the curve's graph.

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 the FTOC helped us find earlier. In order to determine the critical points (max/min), we need:

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

and for the function to be decreasing, the following is required:

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

Evelyn Agard · Answer

undefined To find the intervals where the function $ \int_0^x (t^2 - 6t + 8) dt $ is decreasing, we first need to evaluate the integral and then analyze its derivative.

The integral of the function $ t^2 - 6t + 8 $ with respect to $ t $ is $ \frac{t^3}{3} - 3t^2 + 8t + C $, where $ C $ is the constant of integration.

Taking the derivative of this integral function yields the derivative of the original function $ t^2 - 6t + 8 $.

The derivative of $ \frac{t^3}{3} - 3t^2 + 8t + C $ with respect to $ t $ is $ t^2 - 6t + 8 $.

To find where this derivative is negative (indicating a decreasing function), we solve for the critical points by setting the derivative equal to zero and then test intervals around those points.

$ t^2 - 6t + 8 = 0 $ gives us the critical points $ t = 2 $ and $ t = 4 $.

We can test the intervals $ (-\infty, 2), (2, 4), $ and $ (4, \infty) $ by plugging in test points into the derivative $ t^2 - 6t + 8 $ to see where it is negative.

For example, testing $ t = 0 $ in the derivative gives $ (0)^2 - 6(0) + 8 = 8 $, which is positive, indicating the function is increasing on the interval $ (0, 2) $.

Similarly, testing $ t = 3 $ in the derivative gives $ (3)^2 - 6(3) + 8 = -1 $, which is negative, indicating the function is decreasing on the interval $ (2, 4) $.

Testing a point greater than 4, such as $ t = 5 $, gives $ (5)^2 - 6(5) + 8 = 7 $, which is positive, indicating the function is increasing on the interval $ (4, \infty) $.

Therefore, the function $ \int_0^x (t^2 - 6t + 8) dt $ is decreasing on the interval $ (2, 4) $.