Question: The function f(x) is defined as f(x) = x^3 - 4x^2 + 3x + 2. Find the definite integral of f(x) from x = -1 to x = 3.

Answer: To find the definite integral of f(x) from x = -1 to x = 3, we can use the fundamental theorem of calculus. This states that if F(x) is an antiderivative of f(x), then: ∫[a, b] f(x) dx = F(b) - F(a)

We can start by finding the antiderivative F(x) of f(x) by using the power rule for integration: F(x) = ∫ f(x) dx F(x) = ∫ (x^3 - 4x^2 + 3x + 2) dx F(x) = (1/4)x^4 - (4/3)x^3 + (3/2)x^2 + 2x + C, where C is the constant of integration

Now, we can find the definite integral from x = -1 to x = 3: ∫[-1, 3] f(x) dx = F(3) - F(-1) ∫[-1, 3] f(x) dx = [(1/4)(3^4) - (4/3)(3^3) + (3/2)(3^2) + 23 + C] - [(1/4)(-1^4) - (4/3)(-1^3) + (3/2)(-1^2) + 2(-1) + C] ∫[-1, 3] f(x) dx = [(1/4)(81) - (4/3)(27) + (3/2)(9) + 6 + C] - [(1/4) - (4/3) + (3/2) - 2 + C] ∫[-1, 3] f(x) dx = [20.25 - 36 + 13.5 + 6 + C] - [0.25 + (4/3) - 1.5 - 2 + C] ∫[-1, 3] f(x) dx = 3.75

Therefore, the definite integral of f(x) from x = -1 to x = 3 is 3.75.