Problem 61 - 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. $$ \int_{-2}^{5}|x| d x $$

Step-by-Step Solution

Answer

The integral \( \int_{-2}^{5} |x| \, dx \) evaluates to \( 14.5 \).

Step 1: Understand the Problem

We are tasked with finding the value of the integral \( \int_{-2}^{5}|x| dx \). This integral can be interpreted as the area under the curve of \( y = |x| \) from \( x = -2 \) to \( x = 5 \).

Step 2: Analyze the Function

The function \( y = |x| \) is a piecewise linear function, defined as \( y = -x \) for \( x < 0 \) and \( y = x \) for \( x \geq 0 \). This creates a V-shape with its vertex at the origin \((0,0)\).

Step 3: Divide the Integral into Sections

We divide the integral at \( x = 0 \) due to the piecewise nature of \( |x| \). Thus, we have two integrals: \( \int_{-2}^{0} -x \, dx \) and \( \int_{0}^{5} x \, dx \).

Step 4: Calculate the Area from \(-2\) to \(0\)

The graph from \( x = -2 \) to \( x = 0 \) is a triangle under the line \( y = -x \). The base of this triangle is \( 2 \) units (from \(-2\) to \(0\)) and the height is \( 2 \) (at \( x = -2, \ y = 2 \)). The area of this triangle is \( \frac{1}{2} \times 2 \times 2 = 2 \).

Step 5: Calculate the Area from \(0\) to \(5\)

Here, the graph follows the line \( y = x \), forming a right triangle above the x-axis. The base is \( 5 \) units (from \(0\) to \(5\)) and the height is \( 5 \) (at \( x = 5, \ y = 5 \)). This triangle's area is \( \frac{1}{2} \times 5 \times 5 = 12.5 \).

Step 6: Combine the Areas

Both calculated areas are positive since they represent the absolute area. Therefore, the total signed area under the curve of \( |x| \) from \(-2\) to \(5\) is \( 2 + 12.5 = 14.5 \).

Key Concepts

Piecewise Function

A piecewise function is a type of function that is defined by different expressions for different intervals of its domain. In the context of the integral \( \int_{-2}^{5}|x| \, dx \), the function \( y = |x| \) is a classic example. It is expressed differently based on the value of \( x \):

• For \( x < 0 \), the function is \( y = -x \).
• For \( x \geq 0 \), the function is \( y = x \).

This results in a V-shaped graph with a vertex at the origin \((0,0)\). The change in the function's definition at \( x = 0 \) divides the integral into two separate regions, making it easier to compute the area under each segment independently.

Absolute Value Function

The absolute value function, denoted by \(|x|\), is a fundamental mathematical function that gives the distance of a number \(x\) from zero on the real number line, without considering direction. For any real number \(x\), it is defined as:

• \( |x| = x \) if \( x \geq 0 \)
• \( |x| = -x \) if \( x < 0 \)

In our integral, \(|x|\) transforms negative inputs into their positive counterparts, forming the V-shaped graph. This transformation alters the original linear function \(x\) into two line segments that create different interpretations of the area under the curve when integrated over a specific interval.

Area Under Curve

Finding the area under a curve is one of the primary goals of integration, especially in definite integrals. When working with the integral \(\int_{-2}^{5}|x| \, dx\), we interpret it as the sum of the areas under the line segments defined by the piecewise function \( y = |x| \). To calculate this area:

• For \(-2 \leq x < 0\), the area under the curve \( y = -x \) is represented by a right triangle with a base and height of 2 units, giving an area of \( \frac{1}{2} \times 2 \times 2 = 2 \).
• For \(0 \leq x \leq 5\), the area under the curve \( y = x \) is also a right triangle with a base and height of 5 units, yielding \( \frac{1}{2} \times 5 \times 5 = 12.5 \).

Adding these two areas gives a total area of 14.5, which represents the integral's value.

Geometric Interpretation

The geometric interpretation of an integral provides a visual understanding of the concept of area under a curve. In the given problem, interpreting the integral \( \int_{-2}^{5}|x| \, dx \) geometrically helps in visualizing the regions whose areas need to be calculated.For \( y = |x| \), the transformation into a piecewise function creates two geometric regions:

• A triangle below the x-axis for \( -2 \leq x < 0 \), which appears upside down due to the negative slope \( y = -x \).
• A triangle above the x-axis for \( 0 \leq x \leq 5 \), driven by the positive slope \( y = x \).

The V-shaped appearance of the graph is split precisely at \( x = 0 \), a crucial detail that helps in breaking down complex calculations into manageable geometric shapes. This understanding allows one to confidently add these visualized areas to find the total area, equal to the definite integral's result.