AP Calculus AB Exam Question:

Find the antiderivative of the function f(x) = 3x^2 + 4x + 2.

Solution:

To find the antiderivative of a function, we integrate the given function term by term.

∫(3x^2 + 4x + 2) dx = ∫3x^2 dx + ∫4x dx + ∫2 dx

Using the power rule of integration, the antiderivative of x^n is (1/(n+1)) * x^(n+1), where n ≠ -1, the antiderivative of a constant is the constant times x, and the antiderivative of a constant multiplied by x is the constant times (1/2) * x^2.

So, integrating each term separately:

∫3x^2 dx = (3/(2+1)) * x^(2+1) + C1 = x^3 + C1

∫4x dx = (4/(1+1)) * x^(1+1) + C2 = 2x^2 + C2

∫2 dx = 2x + C3

Thus, the antiderivative of the function f(x) = 3x^2 + 4x + 2 is given by:

∫(3x^2 + 4x + 2) dx = x^3 + 2x^2 + 2x + C

where C = C1 + C2 + C3 is the constant of integration.