(1 point) Evaluate the following integral by interprofting it in terms of areas: \[ \int_{0}^{3}\left(\frac{1}{2} x-1\right) d x \] Value of integral $=$

Step 1 ：The integral of a function can be interpreted as the area under the curve of the function. In this case, the function is a straight line with a slope of 1/2 and a y-intercept of -1. The limits of integration are from 0 to 3. Therefore, the integral represents the area of the trapezoid formed by the x-axis, the y-axis, the line \(y = \frac{1}{2}x - 1\), and the line \(x = 3\).

Step 2 ：The area of a trapezoid is given by the formula \(\frac{1}{2}(\text{base1} + \text{base2})(\text{height})\). In this case, base1 is the y-value of the function at \(x = 0\), base2 is the y-value of the function at \(x = 3\), and the height is the distance between \(x = 0\) and \(x = 3\), which is 3.

Step 3 ：Calculate base1, base2, and height: \(\text{base1} = -1.0\), \(\text{base2} = 0.5\), \(\text{height} = 3\)

Step 4 ：Calculate the area: \(\text{area} = -0.75\)

Step 5 ：The area calculated is negative, which means that the area is below the x-axis. This is consistent with the graph of the function, which is below the x-axis for \(x\) in [0, 3]. Therefore, the value of the integral is -0.75.

Step 6 ：Final Answer: The value of the integral is \(\boxed{-0.75}\)