the graph of the piece - wise linear function f, which has a domain of - 2 ≤ x ≤ 3, is shown in the figure…

Explanation:

Step1: Split the integral by intervals

The function (f(x)) is piece - wise. Split (\int_{-2}^{3}f(x)dx=\int_{-2}^{0}f(x)dx+\int_{0}^{3}f(x)dx).

Step2: Calculate (\int_{-2}^{0}f(x)dx)

For (-2\leq x\leq0), (f(x) = 1). Using the integral formula (\int_{a}^{b}kdx=k(b - a)) (where (k = 1), (a=-2), (b = 0)), we have (\int_{-2}^{0}1dx=1\times(0-( - 2))=2).

Step3: Calculate (\int_{0}^{3}f(x)dx)

The function (y = f(x)) for (0\leq x\leq3) is a line passing through ((0,1)) and ((3,-2)). The equation of the line is (y-1=\frac{-2 - 1}{3-0}(x - 0)), i.e., (y=-x + 1). Using the integral formula (\int_{a}^{b}(-x + 1)dx=\left[-\frac{1}{2}x^{2}+x\right]_{0}^{3}=-\frac{1}{2}(3)^{2}+3=-\frac{9}{2}+3=-\frac{3}{2}).

Step4: Sum the two integrals

(\int_{-2}^{3}f(x)dx=\int_{-2}^{0}f(x)dx+\int_{0}^{3}f(x)dx=2-\frac{3}{2}=\frac{4 - 3}{2}=\frac{1}{2}).

Answer:

A. (\frac{1}{2})