Problem 31 - Calculus Early Transcendentals

Question

Net area and definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand, show the region in question, and interpret your result. $$\int_{0}^{4} f(x) d x, \text { where } f(x)=\left\\{\begin{array}{ll} 5 & \text { if } x \leq 2 \\ 3 x-1 & \text { if } x > 2 \end{array}\right.$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral of the piecewise function f(x), where f(x) = 5 for x ≤ 2 and f(x) = 3x - 1 for x > 2, from x = 0 to x = 4. Answer: The value of the definite integral is 17.

Step 1: Sketch the graph and region

First, we need to sketch the graph of the function f(x). Since it's a piecewise function, we have two parts to sketch. - For x ≤ 2, f(x) = 5, which is a horizontal line at y = 5. - For x > 2, f(x) = 3x - 1, which is a straight line with a slope of 3 and intercepts at (1/3, 0) and (1, 2). Now, as per the given limits of integration (0 to 4), the first part runs from 0 to 2, while the second part runs from 2 to 4. The region to be found is the areas between the graph of f(x) and the x-axis for these intervals.

Step 2: Find the area under each function

For each part of the function, we'll calculate the area of the region. - For x ≤ 2, f(x) = 5: - The area is a rectangle with width 2 (start at 0, end at 2) and height 5. - Area = base * height = 2 * 5 = 10 square units. - For x > 2, f(x) = 3x - 1: - The area is a trapezoid with parallel sides 2 (value at x=2) and 5 (value at x=4) and height 2 (the interval from 2 to 4). - Area = 0.5 * (sum of parallel sides) * height = 0.5 * (2 + 5) * 2 = 7 square units.

Step 3: Add the areas

Now, we need to combine the areas found to get the result of the definite integral. - Total area = Area1 + Area2 = 10 + 7 = 17 square units. Thus, the value of the definite integral is: $$\int_{0}^{4} f(x) d x = 17$$

Key Concepts

Piecewise Functions

Piecewise functions are mathematical functions that have different expressions for different intervals of their input variable. In our exercise, this means that the function behaves differently depending on the value of \( x \). For instance:

• For \( x \leq 2 \), the function \( f(x) = 5 \) is a simple horizontal line at a constant value of 5.
• For \( x > 2 \), the function \( f(x) = 3x - 1 \) is a linear equation with a slope of 3, moving upwards as \( x \) increases.

These kind of functions can model real-world scenarios where a situation changes its behavior at certain points. They are vital in representing systems that have different conditions, like tax brackets or piecewise pricing, where each piece or segment represents a particular condition in the system.
To solve these functions, identify each interval, find the expression related to it, and use these to evaluate or sketch the function, ensuring to take note of changes at the boundaries of each interval.

Area under a Curve

Finding the area under a curve of a given function is a fundamental aspect of calculus, particularly when dealing with definite integrals. In our problem, the task was to find the area under the curve of the piecewise function \( f(x) \) over the interval from \( x = 0 \) to \( x = 4 \).
For piecewise functions, this involves breaking down the total area into smaller, more manageable shapes, such as rectangles and trapezoids, which correspond to each segment of the piecewise function.

• The area under the constant part of the curve (\( f(x) = 5 \) for \( x \leq 2 \)) forms a rectangle.
• The linear segment (\( f(x) = 3x - 1 \) for \( x > 2 \)) forms a trapezoid between \( x = 2 \) and \( x = 4 \).

Breaking the problem into simple geometric shapes makes it easier to calculate and combine these areas to find the total area under the curve. Each area represents a portion of the net area enclosed by the function and the x-axis.

Geometric Interpretation of Integrals

The geometric interpretation of integrals allows us to visually understand how calculus connects to geometry. In essence, the definite integral of a function over an interval gives the net area between the curve of the function and the x-axis.
In our example, the integral \( \int_{0}^{4} f(x) dx \) represents the cumulative area from \( x = 0 \) to \( x = 4 \), combining both positive and negative contributions (if any). Here, it covers:

• A rectangle, from 0 to 2, representing the area under \( f(x) = 5 \) which is above the x-axis.
• A trapezoid, from 2 to 4, representing the area under \( f(x) = 3x - 1 \) also above the x-axis.

This interpretation is crucial as it allows us not only to compute areas and solve practical problems but also to better understand the behavior of different segments of a function. Such interpretations are powerful in physics, engineering, and economics where visualizing the net change can lead to deeper insights and understandings of phenomena.