Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Calculate \(\int_{a}^{b} f(x) d x\), where a and \(b\) are the left and right end points for which f is defined, by using the Interval Additive Property and the appropriate area formulas from plane geometry. Begin by graphing the given function. $$ f(x)=\left\\{\begin{array}{ll} 2 x & \text { if } 0 \leq x \leq 1 \\ 2 & \text { if } 1

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{27}{2}\).

Step 1: Graph the Piecewise Function

Before solving the integral, we will graph the piecewise function to better understand it. Here is the function broken down into its intervals: 1. In the interval \(0 \leq x \leq 1\), the function \(f(x) = 2x\) is a line starting at the origin and going up with a slope of 2.2. In the interval \(1 < x \leq 2\), the function \(f(x) = 2\) is a horizontal line at \(y = 2\).3. In the interval \(2 < x \leq 5\), the function \(f(x) = x\) is a line starting at \(x = 2\) with a slope of 1. Plot these segments on a graph to visualize how they connect.

Step 2: Break Down the Interval for Integration

We will use the Interval Additive Property to integrate the piecewise function by breaking it down into three separate integrals corresponding to the intervals defined by the function:1. \(\int_{0}^{1} 2x \, dx\)2. \(\int_{1}^{2} 2 \, dx\)3. \(\int_{2}^{5} x \, dx\)

Step 3: Calculate Each Integral

Now, compute the integrals over each interval:1. For \(\int_{0}^{1} 2x \, dx\) using the formula for the integral of a power of x, we get: \[\int 2x \, dx = x^2 + C\] Calculate from 0 to 1: \[ [x^2]_{0}^{1} = (1)^2 - (0)^2 = 1\] 2. For \(\int_{1}^{2} 2 \, dx\), since it is a constant function: \[ \int 2 \, dx = 2x + C\] Calculate from 1 to 2: \[ [2x]_{1}^{2} = 2(2) - 2(1) = 4 - 2 = 2\] 3. For \(\int_{2}^{5} x \, dx\): \[ \int x \, dx = \frac{1}{2}x^2 + C\] Calculate from 2 to 5: \[ [\frac{1}{2}x^2]_{2}^{5} = \frac{1}{2}(5)^2 - \frac{1}{2}(2)^2 = \frac{25}{2} - \frac{4}{2} = \frac{21}{2}\]

Step 4: Sum the Areas for the Total Integral

To find the total integral value over the interval \([0, 5]\), sum the results from each of the calculated integrals:\[ 1 + 2 + \frac{21}{2} = \frac{2}{2} + \frac{4}{2} + \frac{21}{2} = \frac{27}{2} \]Therefore, the value of the integral over the entire interval is \(\frac{27}{2}\).

Key Concepts

Piecewise Functions

Understanding piecewise functions is crucial when dealing with real-world applications that feature different rules over distinct intervals. A piecewise function is defined using several subfunctions, where each one applies to a certain interval of the domain.
For example, with the function given in our exercise:

• In the interval \(0 \leq x \leq 1\), it's described by the rule \(f(x) = 2x\). This represents a straight line starting from the origin and rising with a slope of 2.
• In the interval \(1 < x \leq 2\), the rule \(f(x) = 2\) applies. This is a horizontal line, illustrating constant behavior at \(y = 2\).
• Lastly, in the interval \(2 < x \leq 5\), \(f(x) = x\) is used. Here, the line begins at point \(x = 2\) with a slope of 1.

Graphing these segments helps visualize how the function changes across different intervals. It's essential to carefully identify the transition points and the corresponding rules for each segment.
Such functions are particularly useful when modeling systems that change behavior at specified conditions. This could represent systems like a thermostat control or pricing plans that differ based on consumption.

Interval Additive Property

The Interval Additive Property is a fundamental concept when working with integrals, especially with piecewise functions. This property allows us to break down an integral over a complex interval into smaller, more manageable parts. Here's how it works:
Consider the integral of a piecewise function over an interval \([a, b]\). The Interval Additive Property suggests that you can split this interval into smaller sections and integrate over each interval separately:

• \(\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx\)
• Each of these smaller integrals is then added up to achieve the total integral over \([a, b]\).

This logical breakdown allows us to handle each segment of a piecewise function based on its defining rule. In our exercise, the overall integral from 0 to 5 was split into three integrals corresponding to the domain sections of the piecewise-defined \(f(x)\).
This property is highly useful, as it simplifies the task and enhances understanding, especially when each segment represents distinct geometric shapes.

Plane Geometry

When calculating the definite integral of a piecewise function, plane geometry plays a significant role. By relating the areas computed from the integrals with geometric shapes, we can better visualize and understand the solution.
Here are simple geometrical shapes that correspond to each segment in our exercise:

• For \(\int_{0}^{1} 2x \, dx\), imagine the area under the curve as a right triangle with base 1 and height 2.
• The integral \(\int_{1}^{2} 2 \, dx\) relates to a rectangle with length 1 and height 2.
• Finally, \(\int_{2}^{5} x \, dx\) covers a trapezoid, starting from \(y = 2\) extending to \(y = 5\).

Visualizing these shapes helps us to connect the abstract concept of integration with tangible areas. Calculating and summing these areas using basic geometry reinforces why these shapes align with the integral computations in each interval.
This approach is especially beneficial in educational settings, where connecting geometric visualization with analytical solutions promotes a deeper understanding of calculus and its applications in different fields.