14 the graph of the function f shown below consists of two line - segments, as shown in the figure below…

Explanation:

Step1: Split the integral based on intervals

The function (y = f(x)) has different behaviors on ([0,1]) and ([1,3]). So, (\int_{0}^{3}|f(x)|dx=\int_{0}^{1}|f(x)|dx+\int_{1}^{3}|f(x)|dx).

Step2: Find the equation of the first - line segment on ([0,1])

The line segment on ([0,1]) passes through ((0,1)) and ((1,2)). The equation of the line using the point - slope form (y - y_1=m(x - x_1)) with (m=\frac{2 - 1}{1-0}=1) and ((x_1,y_1)=(0,1)) is (y=x + 1). On ([0,1]), (f(x)=x + 1>0), so (\int_{0}^{1}|f(x)|dx=\int_{0}^{1}(x + 1)dx). Using the power rule (\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have (\int_{0}^{1}(x + 1)dx=\left[\frac{x^{2}}{2}+x\right]_{0}^{1}=\frac{1}{2}+1=\frac{3}{2}).

Step3: Find the equation of the second - line segment on ([1,3])

The line segment on ([1,3]) passes through ((1,1)) and ((3,-1)). The slope (m=\frac{-1 - 1}{3 - 1}=-1). Using the point - slope form with ((x_1,y_1)=(1,1)), the equation is (y-1=-(x - 1)), or (y=-x + 2). On ([1,2]), (f(x)=-x + 2\geq0), and on ([2,3]), (f(x)=-x + 2<0). (\int_{1}^{3}|f(x)|dx=\int_{1}^{2}(-x + 2)dx+\int_{2}^{3}-(-x + 2)dx).

• Calculate (\int_{1}^{2}(-x + 2)dx=\left[-\frac{x^{2}}{2}+2x\right]_{1}^{2}=(-2 + 4)-\left(-\frac{1}{2}+2\right)=2-\frac{3}{2}=\frac{1}{2}).
• Calculate (\int_{2}^{3}-(-x + 2)dx=\int_{2}^{3}(x - 2)dx=\left[\frac{x^{2}}{2}-2x\right]{2}^{3}=\left(\frac{9}{2}-6\right)-(2 - 4)=\frac{9 - 12}{2}+2=\frac{-3 + 4}{2}=\frac{1}{2}). So, (\int{1}^{3}|f(x)|dx=\frac{1}{2}+\frac{1}{2}=1).

Step4: Calculate the total integral

(\int_{0}^{3}|f(x)|dx=\int_{0}^{1}|f(x)|dx+\int_{1}^{3}|f(x)|dx=\frac{3}{2}+1=\frac{5}{2}).

Answer:

A. (\frac{5}{2})