Answer is: option2

Solution:

We are given the expression:

\[ \int_{-3}^{3} f(x) \, dx - 2 \int_{-1}^{3} f(x) \, dx \]

We are told:

1. The area of the region from \(-3\) to \(-1\), where \(f(x) > 0\), is \(A\).
2. The area from \(-1\) to \(3\), where \(f(x) < 0\), is \(B\).

Since integrals take into account the sign of the function:

1. \[ \int_{-3}^{-1} f(x) \, dx = A \] because \(f(x) > 0\)
2. \[ \int_{-1}^{3} f(x) \, dx = -B \] because \(f(x) < 0\)

So:

\[ \int_{-3}^{3} f(x) \, dx = \int_{-3}^{-1} f(x) \, dx + \int_{-1}^{3} f(x) \, dx = A + (-B) = A - B \]

Now compute the full expression:

\[ \int_{-3}^{3} f(x) \, dx - 2 \int_{-1}^{3} f(x) \, dx = (A - B) - 2(-B) = A - B + 2B = A + B \]

Final Answer: (B) \(A + B\)