The graph of \( f \) is shown in the figure. Let \( A(x)=\int_{-2}^{x} f(t) d t \) and \( F(x)=\int_{4}^{x} f(t) d t \) be two area functions for \( f \). Evaluate the following area functions. a. \( A(-2)=\square \) (Simplify your answer.)

To evaluate \( A(-2) \), we need to look at the definition of the area function: \[ A(x) = \int_{-2}^{x} f(t) \, dt. \] When we set \( x = -2 \), the integral becomes: \[ A(-2) = \int_{-2}^{-2} f(t) \, dt. \] The integral of any function over an interval where the upper and lower limits are the same (in this case, both are \(-2\)) is zero. Hence, we have: \[ A(-2) = 0. \] So the final answer is: \( A(-2) = 0 \).