Question: Consider the function f(x) = 2x^3 - 5x^2 + 3x + 1. This represents the position of an object moving along the x-axis at time t. The derivative of this function, f'(x), represents the velocity of the object.
a) Find the net change of the object's position from x = 1 to x = 5. b) Determine the position at t = 0, and find the total accumulation of the object's position from t = 0 to t = 3.
Answer:
a) To find the net change of the object's position from x = 1 to x = 5, we need to calculate the definite integral of f'(x) from x = 1 to x = 5.
First, let's find f'(x), which represents the velocity of the object: f'(x) = d/dx(2x^3 - 5x^2 + 3x + 1) = 6x^2 - 10x + 3
Now, we can calculate the definite integral of f'(x) from x = 1 to x = 5: Net change = ∫(a to b) f'(x) dx = ∫(1 to 5) (6x^2 - 10x + 3) dx
Taking the anti-derivative of each term: Net change = [2x^3 - 5x^2 + 3x] (1 to 5) = [(2(5)^3 - 5(5)^2 + 3(5)) - (2(1)^3 - 5(1)^2 + 3(1))]
Simplifying: Net change = [(250 - 125 + 15) - (2 - 5 + 3)] = [140 - 0] = 140 units (Answer)
Therefore, the net change of the object's position from x = 1 to x = 5 is 140 units.
b) To find the position at t = 0, we need to evaluate f(0), which will give us the initial position of the object.
f(x) = 2x^3 - 5x^2 + 3x + 1 f(0) = 2(0)^3 - 5(0)^2 + 3(0) + 1 = 0 - 0 + 0 + 1 = 1
Therefore, the position at t = 0 is 1 unit.
To find the total accumulation of the object's position from t = 0 to t = 3, we need to calculate the definite integral of f(x) from t = 0 to t = 3.
Total Accumulation = ∫(a to b) f(x) dx = ∫(0 to 3) (2x^3 - 5x^2 + 3x + 1) dx
Taking the anti-derivative of each term: Total Accumulation = [∫ 2x^3 dx - ∫ 5x^2 dx + ∫ 3x dx + ∫ 1 dx] (0 to 3) = [(2/4)x^4 - (5/3)x^3 + (3/2)x^2 + x] (0 to 3)
Substituting x = 3: Total Accumulation = [(2/4)(3)^4 - (5/3)(3)^3 + (3/2)(3)^2 + 3] - [(2/4)(0)^4 - (5/3)(0)^3 + (3/2)(0)^2 + 0]
Simplifying: Total Accumulation = [(2/4)(81) - (5/3)(27) + (3/2)(9) + 3] - [0 - 0 + 0 + 0] = [40.5 - 45 + 13.5 + 3] - [0] = 52 units (Answer)
Therefore, the total accumulation of the object's position from t = 0 to t = 3 is 52 units.