AP Calculus AB Exam Question:

Let f(x) be a continuous function on the closed interval [a, b] and let F(x) be the antiderivative of f(x) on the interval [a, b]. The Fundamental Theorem of Calculus states that if the integral of f(x) from a to x is defined as:

I(x) = ∫[a,x] f(t) dt

where a ≤ x ≤ b, then the derivative of F(x) with respect to x is equal to f(x), or in symbols:

d/dx [F(x)] = f(x)

Given that f(x) = 2x^3 - 4x and F(x) = x^4 - 2x^2 + C, find the definite integral of f(x) from -1 to 2 using the Fundamental Theorem of Calculus.

Answer:

The Fundamental Theorem of Calculus states that if F(x) is the antiderivative of f(x), then the definite integral of f(x) from a to b can be evaluated by subtracting the antiderivative values at a and b:

∫[a,b] f(x) dx = F(b) - F(a)

In this case, we are given f(x) = 2x^3 - 4x and F(x) = x^4 - 2x^2 + C.

To evaluate the definite integral of f(x) from -1 to 2, we substitute the limits of integration into the antiderivative F(x):

∫[-1,2] (2x^3 - 4x) dx = F(2) - F(-1)

Substituting x = 2 into F(x):

F(2) = (2)^4 - 2(2)^2 + C = 16 - 8 + C = 8 + C

Next, substituting x = -1 into F(x):

F(-1) = (-1)^4 - 2(-1)^2 + C = 1 - 2 + C = -1 + C

Now we can calculate the definite integral:

∫[-1,2] (2x^3 - 4x) dx = F(2) - F(-1) = (8 + C) - (-1 + C) = 8 - (-1) + C - C = 9

Therefore, the value of the definite integral of f(x) from -1 to 2 is 9.