Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-22, 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)= \begin{cases}2 x & \text { if } 0 \leq x \leq 1 \\ 2 & \text { if } 1

Step-by-Step Solution

Answer

The integral of the function over the given range is 13.5.

Step 1: Graph the Function

To understand the function better, graph it for the given intervals. From \(0\) to \(1\), the function is \(2x\). The graph will be a line starting from (0,0) and going to (1,2). From \(1\) to \(2\), the function is constant at \(2\). This appears as a horizontal line from (1,2) to (2,2). From \(2\) to \(5\), the function is \(x\). This is a line starting from (2,2) and extending to (5,5).

Step 2: Integrate \(f(x) = 2x\) from 0 to 1

For the interval \([0, 1]\), the function is \(f(x) = 2x\). The area under this line (a triangle) is found using the area formula for a triangle: \(\frac{1}{2} \times \text{base} \times \text{height}\). The base is \(1\) and the height is \(2\), so the area is \(\frac{1}{2} \times 1 \times 2 = 1\).

Step 3: Integrate \(f(x) = 2\) from 1 to 2

For the interval \([1, 2]\), the function is \(f(x) = 2\), a horizontal line. Thus, it forms a rectangle with a height of \(2\) and a width of \(1\). The area of the rectangle is \(\text{width} \times \text{height} = 1 \times 2 = 2\).

Step 4: Integrate \(f(x) = x\) from 2 to 5

For the interval \([2, 5]\), the function is \(f(x) = x\). This forms a trapezoid when considering the area between \(x=2\) and \(x=5\). The height of the trapezoid is \(5 - 2 = 3\), the base lengths are the functional values at \(x=2\) and \(x=5\), which are \(2\) and \(5\) respectively. The area formula for the trapezoid is \(\frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}\), so \(\frac{1}{2} \times (2 + 5) \times 3 = \frac{21}{2} = 10.5\).

Step 5: Sum the Areas

Add up the areas from each interval: area from \([0,1]\) is \(1\), from \([1,2]\) is \(2\), and from \([2,5]\) is \(10.5\). Therefore, the total area is \(1 + 2 + 10.5 = 13.5\).

Key Concepts

Piecewise Functions

Piecewise functions are unique in that they consist of different expressions for different intervals of the input variable. This allows them to model various real-world scenarios where behavior changes over a range. In the given exercise, the function \(f(x)\) switches at specific points:

• From \(x = 0\) to \(x = 1\), it's defined as \(2x\), forming a line segment.
• From \(x = 1\) to \(x = 2\), it becomes constant at \(f(x) = 2\).
• From \(x = 2\) to \(x = 5\), it changes again to \(f(x) = x\).

These intervals reflect how the function behaves differently across the domain. To solve integrals involving piecewise functions, it's essential to break calculations into the specific sections where these different expressions are valid.

Area Under the Curve

Finding the area under a curve represented by a function is a core concept in calculus. This area represents a definite integral over a specific interval and has numerous practical applications such as in physics, statistics, and economics. For a piecewise function like \(f(x)\), we calculate each segment's area separately using different geometric formulas:

• For the segment \(f(x) = 2x\), between \(0\) and \(1\), the area is a triangle.
• For \(f(x) = 2\), from \(1\) to \(2\), the shape is a rectangle.
• For \(f(x) = x\), between \(2\) and \(5\), a trapezoid is formed.

Each geometric shape corresponds to an interval and gives the specific area under the function's curve for that interval.

Interval Additive Property

The interval additive property of integrals plays a crucial role in evaluating the definite integral of piecewise functions. It states that the integral over a disjoint sum of intervals can be computed by separately integrating over each interval and then summing up the results. For the function in the exercise, this property is applied as follows:

• Calculate the integral from \(0\) to \(1\), \([\int_{0}^{1} 2x \, dx]\).
• Then from \(1\) to \(2\), \([\int_{1}^{2} 2 \, dx]\).
• Finally from \(2\) to \(5\), \([\int_{2}^{5} x \, dx]\).

By adding up these individual integrals, we find the total area under the curve across the entire domain of the function, which is essential for solving the given problem efficiently.

Graphical Representation

Graphically representing piecewise functions helps in visualizing and understanding the function's behavior over different intervals. By plotting \(f(x)\), you can observe:

• A line segment from \((0,0)\) to \((1,2)\) representing \(2x\).
• A horizontal line from \((1,2)\) to \((2,2)\) for the constant part \(f(x) = 2\).
• A line from \((2,2)\) to \((5,5)\) depicting \(f(x) = x\).

This visual aid not only supports the mathematical analysis but also aids in identifying the types of geometric areas involved—triangles, rectangles, and trapezoids. Therefore, graphing is a strong strategy to clarify the different segments of a piecewise function before integrating.