Problem 83 - Calculus Volume 1

Question

In the following exercises, evaluate the integral using area formulas. \(\int_{-2}^{3}(3-|x|) d x\)

Step-by-Step Solution

Answer

The integral evaluates to 7.5.

Step 1: Understand the Function

The integrand is given by the function \( f(x) = 3 - |x| \), which is a piecewise linear function. Understanding the function's behavior is crucial: for \( x < 0 \), \( |x| = -x \) resulting in \( f(x) = 3 + x \), and for \( x \geq 0 \), \( |x| = x \) leading to \( f(x) = 3 - x \).

Step 2: Visualize the Graph

The function \( f(x) = 3 + x \) when \( x < 0 \) is a line with a positive slope, and \( f(x) = 3 - x \) when \( x \ge 0 \) is a line with a negative slope. The graph is a V-shaped figure centered at the origin with peaks at (0,3) and intercepts at (-3,0) and (3,0).

Step 3: Split Integral Based on Intervals

The integration limits span \([-2, 3]\). Split the integral at \( x = 0 \), where the absolute function changes: \[ \int_{-2}^{3} (3 - |x|) \, dx = \int_{-2}^{0} (3 + x) \, dx + \int_{0}^{3} (3 - x) \, dx \].

Step 4: Apply Area Formula for First Interval

Calculate the area under \( f(x) = 3 + x \) from \( x = -2 \) to \( x = 0 \). This forms a right triangle with base 2 and height 3. The area is given by \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 3 = 3 \].

Step 5: Apply Area Formula for Second Interval

Calculate the area under \( f(x) = 3 - x \) from \( x = 0 \) to \( x = 3 \). This also forms a right triangle with base 3 and height 3. The area is \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 3 = 4.5 \].

Step 6: Combine Areas

Add the areas obtained from both regions. The total area under the curve from \( x = -2 \) to \( x = 3 \) is \[ 3 + 4.5 = 7.5 \].

Key Concepts

Piecewise Functions

Piecewise functions are special because they are defined by different expressions depending on the input value. This means that a single function can switch behaviors at specified points, creating a more complex graph. In the exercise given, the function is represented as two distinct linear segments:

• For values when \( x < 0 \), the expression is \( f(x) = 3 + x \).
• For values when \( x \geq 0 \), the expression is \( f(x) = 3 - x \).

This kind of function is useful for modeling situations where a sudden change occurs, such as crossing over to a new tax bracket or different rate. The visual representation often forms separate curves or lines pieced together, such as the V-shape in the step-by-step solution, where it changes at the critical point \( x = 0 \). Understanding the form of these functions helps predict the overall behavior based on the individual segments.

Area Under a Curve

Calculating the area under a curve is a fundamental concept in integral calculus, often used to determine total quantities, such as distance, mass, or probability. In this exercise, the curve is defined by the piecewise linear function \( 3 - |x| \), and the task is to find the total area between the curve and the x-axis from \( x = -2 \) to \( x = 3 \).When dealing with piecewise functions, you must first split the graph based on the segments of the piecewise function. In this case:

• From \( x = -2 \) to \( x = 0 \), you form one triangular area under \( f(x) = 3 + x \).
• From \( x = 0 \) to \( x = 3 \), you form another triangular area under \( f(x) = 3 - x \).

By calculating these separate areas and then combining them, you can find the total area under the original piecewise function.

Definite Integrals

Definite integrals represent the exact area under a curve between two points. They are essential in solving problems where you need the cumulative value, such as total distance covered or volume. In this exercise, the definite integral is expressed as:\[ \int_{-2}^{3}(3-|x|) \, dx \]When solved, this provides the total area under the function from \( x = -2 \) to \( x = 3 \). Key steps include:

• Determination of where the piecewise definition changes, in this example at \( x = 0 \).
• Evaluation of the integral separately for each segment: \( \int_{-2}^{0}(3+x) \, dx \) and \( \int_{0}^{3}(3-x) \, dx \).
• Summation of these areas to find the total area, which is 7.5 in this case.

Understanding definite integrals involves recognizing the limits of integration, which dictate the boundaries of the area being calculated. By evaluating a definite integral, we can accurately determine the total quantity represented by the piecewise function within those limits.