Problem 63 - Calculus for Biology and Medicine

Question

Use an area formula from geometry to find the value of each integral by interpreting it as the (signed) area under the graph of an appropriately chosen function. $$ \int_{2}^{5}\left(\frac{1}{2} x-4\right) d x $$

Step-by-Step Solution

Answer

The value of the integral is \(-6.75\).

Step 1: Identify the Function and Interval

The given integral is \( \int_{2}^{5} \left( \frac{1}{2}x - 4 \right) \, dx \). It represents the area under the graph of the linear function \( f(x) = \frac{1}{2}x - 4 \) over the interval \([2, 5]\).

Step 2: Sketch the Graph and Locate Points

The function \( f(x) = \frac{1}{2}x - 4 \) is a straight line with a slope of \( \frac{1}{2} \) and a y-intercept at \(-4\). Calculate the function's values at the endpoints of the interval:\begin{align*}f(2) & = \frac{1}{2}(2) - 4 = -3 \f(5) & = \frac{1}{2}(5) - 4 = -1.5\end{align*}Plot these points: (2, -3) and (5, -1.5).

Step 3: Determine Area Under the Curve

Since the function is linear and the graph forms a trapezoid between \(x=2\) and \(x=5\), use the formula for the area of a trapezoid: \[ A = \frac{1}{2} \times (b_1 + b_2) \times h \]Here, \(b_1 = -3\), \(b_2 = -1.5\), and \(h = 5 - 2 = 3\).

Step 4: Calculate the Area

Substitute the values into the trapezoid area formula:\[ A = \frac{1}{2} \times (-3 + (-1.5)) \times 3 = \frac{1}{2} \times (-4.5) \times 3 = -\frac{13.5}{2} = -6.75 \]This result is negative, indicating that the graph lies below the x-axis over the interval, so the signed area (and thus the integral value) is \(-6.75\).

Key Concepts

Area Under Curve

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 \(x = 5\).

Linear Function

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.

Trapezoid Area Formula

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.