Problem 35 - Calculus

Question

Use the Interval Additive Property and linearity to evaluate \(\int_{0}^{4} f(x) d x .\) Begin by drawing a graph off. $$ f(x)=|x-2| $$

Step-by-Step Solution

Answer

The integral evaluates to 4.

Step 1: Understanding the Function

The function given is \( f(x) = |x - 2| \). This is an absolute value function centered at \( x = 2 \). The graph of \( |x - 2| \) is a V-shape with its vertex at the point \( (2,0) \). For values of \( x < 2 \), \( f(x) = 2 - x \), and for \( x > 2 \), \( f(x) = x - 2 \).

Step 2: Using Interval Additive Property

To evaluate \( \int_{0}^{4} f(x) \, dx \), we'll break it into two intervals where the function changes its formula: \( \int_{0}^{2} f(x) \, dx \) and \( \int_{2}^{4} f(x) \, dx \). These correspond to \( f(x) = 2 - x \) for \([0, 2]\) and \( f(x) = x - 2 \) for \([2, 4]\). So, the integral becomes \( \int_{0}^{2} (2-x) \, dx + \int_{2}^{4} (x-2) \, dx \).

Step 3: Evaluating the First Integral

Calculate \( \int_{0}^{2} (2-x) \, dx \):\[= \int_{0}^{2} 2 \, dx - \int_{0}^{2} x \, dx\]\[= \left[ 2x \right]_{0}^{2} - \left[ \frac{x^2}{2} \right]_{0}^{2}\]\[= (4 - 0) - \left( \frac{4}{2} - 0 \right)\]\[= 4 - 2 = 2\].

Step 4: Evaluating the Second Integral

Calculate \( \int_{2}^{4} (x-2) \, dx \):\[= \int_{2}^{4} x \, dx - \int_{2}^{4} 2 \, dx\]\[= \left[ \frac{x^2}{2} \right]_{2}^{4} - \left[ 2x \right]_{2}^{4}\]\[= \left( \frac{16}{2} - \frac{4}{2} \right) - (8 - 4)\]\[= (8 - 2) - 4 = 2\].

Step 5: Adding the Results

Combine the results of the two integrals using the linearity property of integrals:\[\int_{0}^{4} f(x) \, dx = \int_{0}^{2} (2-x) \, dx + \int_{2}^{4} (x-2) \, dx = 2 + 2 = 4\].

Key Concepts

Absolute Value Function

Absolute value functions are instrumental in many mathematical problems. An absolute value function expressed as \( f(x) = |x - a| \) takes the form of a V-shaped graph centered at the point \( x = a \). In this case, the function \( f(x) = |x - 2| \) is centered at \( x = 2 \), and graphically, it forms a V-shape with its vertex at the coordinate (2,0).

The behavior of the absolute value function changes based on its input's location relative to the center. For inputs less than the center, the function gives \( f(x) = a - x \). For inputs greater than or equal to the center: \( f(x) = x - a \). This change in behavior is crucial for evaluating integrals over the function, as different formulas need to be applied across different intervals.

Interval Additive Property

The interval additive property is a fundamental tool in solving integrals, especially for piecewise or segmented functions like the absolute value function.

This property allows us to split a single integral across a broader range into multiple integrals, each covering parts of the range with potentially different characteristics. For example, if you're given an integral from 0 to 4 for the function \( f(x) = |x-2| \), you can break it into two parts:

• \( \int_{0}^{2} f(x) \, dx \)
• \( \int_{2}^{4} f(x) \, dx \)

This division aligns with the points where the function changes formula, making it easier to calculate the area under the curve using simpler functions for each segment.

Linearity of Integrals

Linearity of integrals refers to the property that allows breaking down and simplifying the computation of definite integrals. It is applicable in the process of integrating functions, as it lets us evaluate the sum of integrals separately.

This linearity means:

• You can separate the sum of different functions' integrals over the same interval.
• You can factor out constants from each integral.
  

In practice, when you have a composite function like \( f(x) = |x-2| \), you can use its linearity to handle each piece as separate terms. Given the integrals from 0 to 2 and 2 to 4, you evaluate them independently as:

• \( \int (2-x) \, dx \)
• \( \int (x-2) \, dx \)

By evaluating each independently, you add their results to yield the total computation desired, exploiting the integral's distributive style in addition.