Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 33-36, use the Interval Additive Property and linearity to evaluate \(\int_{0}^{4} f(x) d x\). Begin by drawing a graph of \(f\). $$ f(x)=3+|x-3| $$

Step-by-Step Solution

Answer

The value of the integral \( \int_{0}^{4} f(x) \, dx \) is 17.

Step 1: Understand the function

The function given is \( f(x) = 3 + |x - 3| \). This is a piecewise function and its graph consists of two linear parts. For \( x < 3 \), \( f(x) = 3 + (3-x) = 6 - x \). For \( x \geq 3 \), \( f(x) = 3 + (x-3) = x \). These form two lines meeting at the point \( x = 3 \).

Step 2: Draw the graph of the function

Draw the two linear pieces on the graph. The first piece for \( x < 3 \) is a line from the point (0, 6) to (3, 3) with a negative slope of -1. The second piece for \( x \geq 3 \) is a line starting at (3, 3) and moving to (4, 4) with a positive slope of +1.

Step 3: Split the integral according to the graph

To evaluate \( \int_{0}^{4} f(x) \, dx \) using the Interval Additive Property, split the integral into two parts at the point where the expression for the function changes, which is at \( x = 3 \). Thus, \( \int_{0}^{4} f(x) \, dx = \int_{0}^{3} (6-x) \, dx + \int_{3}^{4} x \, dx \).

Step 4: Evaluate the first integral

Calculate \( \int_{0}^{3} (6-x) \, dx \). The antiderivative of \( 6 - x \) is \( 6x - \frac{x^2}{2} \). Evaluate this from \( 0 \) to \( 3 \): \[ \left[ 6x - \frac{x^2}{2} \right]_{0}^{3} = \left( 6(3) - \frac{3^2}{2} \right) - \left( 6(0) - \frac{0^2}{2} \right) = 18 - 4.5 = 13.5 \]

Step 5: Evaluate the second integral

Calculate \( \int_{3}^{4} x \, dx \). The antiderivative of \( x \) is \( \frac{x^2}{2} \). Evaluate this from \( 3 \) to \( 4 \): \[ \left[ \frac{x^2}{2} \right]_{3}^{4} = \left( \frac{4^2}{2} \right) - \left( \frac{3^2}{2} \right) = 8 - 4.5 = 3.5 \]

Step 6: Sum the results of both integrals

Add the results of the two integrals obtained: \( 13.5 + 3.5 = 17 \).

Key Concepts

Piecewise Functions

A piecewise function is a function composed of multiple segments or pieces, each defined on a certain interval of the domain. In this exercise, the given function is \( f(x) = 3 + |x - 3| \), which can be rewritten as two separate expressions based on the value of \( x \). This function exhibits different behavior depending on whether \( x \) is less than or greater than 3.

For \( x < 3 \), the function is defined as \( f(x) = 6 - x \) — a linear equation with a negative slope. For \( x \geq 3 \), the function changes to \( f(x) = x \), which is a linear equation with a slope of 1. These two linear sections meet at the point \( x = 3 \), creating a V-shaped graph due to the absolute value.

Understanding piecewise functions is crucial because they often model real-world scenarios where a function's rule varies depending on the input, such as tax brackets, shipping costs, or changes in material properties.

Interval Additive Property

The interval additive property of integrals is a powerful concept in calculus. It allows you to split an integral over a larger interval into the sum of integrals over smaller intervals. This property is particularly useful when dealing with piecewise functions, where different expressions define different parts of the function.

In this exercise, the integral \( \int_{0}^{4} f(x) \, dx \) is split into two parts because the piecewise function changes at \( x = 3 \).

• The integral from 0 to 3 of the function \( 6 - x \)
• The integral from 3 to 4 of the function \( x \)

By evaluating these smaller integrals separately, you can handle each piece of the function according to its own rule before combining them to find the overall area under the curve.

Linearity of Integrals

Linearity of integrals is a mathematical property that enables you to simplify calculations involving integrals. It tells us that the integral of a sum of functions is the same as the sum of the integrals of each function. This property also allows us to factor out constants from the integral.

In simple terms:

• If you have \( \int (f(x) + g(x)) \, dx \), this is the same as \( \int f(x) \, dx + \int g(x) \, dx \).
• If there is a constant \( c \), \( \int c \cdot f(x) \, dx \) becomes \( c \cdot \int f(x) \, dx \).

In our exercise, linearity helps simplify the calculations involved in integrating each piece of the function separately. This makes the process more straightforward and manageable, especially when working with complex piecewise functions.

Definite Integrals

Definite integrals are used to compute the area under a curve over a specified interval on the x-axis. Unlike indefinite integrals, which represent a general form of antiderivative, definite integrals calculate a specific numerical value.

For this exercise, evaluating the definite integrals of each piece of the function between specified limits provided the area under the curve from 0 to 4. Start by finding the antiderivative for each segment and then apply the limits of integration.

The definite integral \( \int_{a}^{b} f(x) \, dx \) is evaluated as follows:

• Find the antiderivative \( F(x) \) of \( f(x) \)
• Compute \( F(b) - F(a) \)

This concept is crucial in calculus, as it applies to problems in physics, engineering, economics, and beyond, where determining area or accumulation is required.