91Ó°ÊÓ

Understanding the idea of finding an area under a curve is essential in calculus, especially when dealing with integrals. This area represents the total sum of values that a function outputs over a given interval.
In simpler terms, imagine a function as a unique, picturesque landscape. The area we seek is like laying a giant transparent mat directly below or above this landscape between two points of interest. In practical scenarios:

• If the function is completely above the x-axis, the integral will give you the exact area beneath the curve.
• If the function dips below the x-axis, as in our example, the area computed will be negative. This is referred to as a "signed area," indicating the direction of the area in relation to the x-axis.

Thus, when given an integral such as \( \int_{2}^{5} \left( \frac{1}{2}x - 4 \right) \, dx \), the central task is to compute this area as defined between the curve and the x-axis from \(x = 2\) to \(x = 5\).

A linear function is one of the simplest forms of functions that map input values to output values, typically in the form of a straight line when graphed. The general form of a linear function can be written as \( f(x) = mx + c \), where:

• \(m\) is the slope of the line, showing how steep it is.
• \(c\) is the y-intercept, indicating where the line crosses the y-axis.

For the given function \( f(x) = \frac{1}{2}x - 4 \):

• The slope \( \frac{1}{2} \) means that for every unit increase in \(x\), \(f(x)\) increases by \(0.5\).
• The y-intercept at \(-4\) tells us that the line crosses the y-axis at the point (0, -4).

In this context, each value of \(f(x)\) represents the height of the line at that \(x\) value. By graphing it over a specific interval, such as from \(x = 2\) to \(x = 5\), you can visualize and compute the area under this line.

The trapezoid area formula is a handy tool for finding the area under a linear function, particularly when the line spans across two unequal points on the y-axis. This is how it works:Let’s visualize the interval between \(x = 2\) and \(x = 5\) on the graph of \( f(x) = \frac{1}{2}x - 4 \). The line creates a geometric shape resembling a trapezoid between these points and the x-axis.
The formula to find the area of a trapezoid is:\[A = \frac{1}{2} \times (b_1 + b_2) \times h \]Where:

• \(b_1\) and \(b_2\) are the lengths of the two parallel sides of the trapezoid (here, these are \(f(2)\) and \(f(5)\)).
• \(h\) stands for the height, or the distance between these parallel lines along the x-axis, given by \(5 - 2 = 3\).

Calculating using our values:\[A = \frac{1}{2} \times (-3 + (-1.5)) \times 3 = -6.75\]The negative result is due to the graph being below the x-axis, which gives us the signed area, reflecting the integral's value, \(-6.75\). This method not only provides the result but visually reinforces understanding through shapes.