Problem 18 - Calculus

Question

Calculate \(\int_{a}^{b} f(x) d x,\) where \(a\) and \(b\) are the left and right end points for which fis 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} 3 x & \text { if } 0 \leq x \leq 1 \\ 2(x-1)+2 & \text { if } 1

Step-by-Step Solution

Answer

\( \int_{0}^{2} f(x) \, dx = \frac{9}{2} \).

Step 1: Understanding the Function

The function \( f(x) \) is piecewise-defined. For \( 0 \leq x \leq 1 \), \( f(x) = 3x \) is a linear function forming a straight line through the origin with a slope of 3. For \( 1 < x \leq 2 \), \( f(x) = 2(x-1) + 2 \) simplifies to \( 2x \), which is another linear function starting from \( (1, 2) \) and proceeding with a slope of 2.

Step 2: Setting Interval Bounds

We need to calculate the integral from \( a = 0 \) to \( b = 2 \). The piecewise function spans over two intervals: \([0, 1]\) and \((1, 2]\), and each segment can be evaluated separately.

Step 3: Integrate the First Function Piece

For \( 0 \leq x \leq 1 \), find the area under \( f(x) = 3x \). This is a right-triangle with vertices at \((0,0)\), \((1,0)\), and \((1,3)\). The area is calculated as: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 3 = \frac{3}{2} \).

Step 4: Integrate the Second Function Piece

For \( 1 < x \leq 2 \), evaluate \( f(x) = 2x \). The area under this curve is another right-triangle, with vertices at \((1,2)\), \((2,0)\), and \((2,4)\). The calculated area is \( \text{Area} = \frac{1}{2} \times 1 \times 2 + \text{rectangle} \), where the rectangle is \( 2 \times 1 \). The total area in this section is \( 1 + 2 = 3 \).

Step 5: Calculate Total Area using the Additive Property

By the additive property of integrals, the total area from \( x = 0 \) to \( x = 2 \) is the sum of the areas from each interval. Sum: \( \int_{0}^{1} 3x \, dx = \frac{3}{2} \), and \( \int_{1}^{2} 2x \, dx = 3 \). Therefore, \( \int_{0}^{2} f(x) \, dx = \frac{3}{2} + 3 = \frac{9}{2} \).

Key Concepts

Piecewise Functions

When dealing with piecewise functions, it's important to understand that these are functions defined by multiple sub-functions, each applied to a different interval of the domain. This means that for different values of the input, you might be using different formulas to compute the output. For the function given in our exercise, we have two different expressions:

• When the input, or x, is between 0 and 1 (including 0 and 1), the function is defined as f(x) = 3x.
• When x is between 1 and 2 (including 2 but not 1), the function is defined as f(x) = 2x after simplifying the expression 2(x-1) + 2.

To properly evaluate and graph piecewise functions, follow these steps: - **Identify the intervals**: Determine the ranges for which each piece of the function is applicable. Here, the intervals are [0, 1] and (1, 2]. - **Graph each piece separately**: Plot these graphs only over their specified intervals. This allows you to visualize the complete function, which can be useful for understanding and performing integrations.

Interval Additive Property

The Interval Additive Property is a fundamental concept in calculus, particularly when finding the integral of a piecewise function. This property states that the integral over an interval can be broken down into the sum of integrals over individual sub-intervals. This is essential for piecewise functions, as each section is defined by its own formula.
In practice:

• Identify the distinct intervals as defined by the piecewise function. In the original exercise, these are [0, 1] and (1, 2].
• Compute the integral over each interval separately, using the specific formula for the function segment within that range.
• Finally, sum up the individual integrals to find the total area under the curve across the entire initial interval, which was [0, 2] in this case.

This step-by-step accumulation of areas respects how the segments contribute to the whole, making it possible to handle functions that aren't uniform across their range.

Area Under a Curve

The area under a curve provides a visual representation of the definite integral of a function over a specific interval. Calculating this area is crucial for understanding how the value of the function changes across that range. For the given piecewise function:

• The first segment, f(x) = 3x, over the interval [0, 1], describes a line forming a right triangle. The area of this triangle is calculated using the formula for the area of a triangle (\[\frac{1}{2} \times \text{base} \times \text{height}\]) which gives an area of \(\frac{3}{2}\).
• The second segment, f(x) = 2x, from (1, 2] is another right triangle (and a rectangle), together evaluated to yield an area of 3.

These specific calculations are linked to geometric shapes due to the linear nature of the function in specified intervals. By summing the computed areas, we obtain the total area under the piecewise function from 0 to 2, resulting in \(\frac{9}{2}\). This visualization aids not only in understanding the numerical answer but also in grasping the geometrical interpretation of integrals.