Problem 9 - University Calculus: Early Transcendentals

Question

Suppose that \(f\) and \(g\) are integrable and that $$\int_{1}^{2} f(x) d x=-4, \quad \int_{1}^{5} f(x) d x=6, \quad \int_{1}^{5} g(x) d x=8$$ Use the rules in Table 5.6 to find a. \(\int_{2}^{2} g(x) d x\) b. \(\int_{5}^{1} g(x) d x\) c. \(\int_{1}^{2} 3 f(x) d x\) d. \(\int_{2}^{5} f(x) d x\) e. \(\int_{1}^{5}[f(x)-g(x)] d x\) f. \(\int_{1}^{5}[4 f(x)-g(x)] d x\)

Step-by-Step Solution

Answer

a) 0, b) -8, c) -12, d) 10, e) -2, f) 16.

Step 1: Understand Property of Definite Integrals

The definite integral \( \int_{a}^{a} g(x) \, dx \) for any function \( g(x) \) is always zero because the upper and lower limits of integration are the same.

Step 2: Apply Property to Solve Part (a)

Using the property of integrals that \( \int_{a}^{a} g(x) \, dx = 0 \), we find that \( \int_{2}^{2} g(x) \, dx = 0 \).

Step 3: Evaluate Integral in Reverse for Part (b)

The property \( \int_{a}^{b} g(x) \, dx = -\int_{b}^{a} g(x) \, dx \) is used. Given \( \int_{1}^{5} g(x) \, dx = 8 \), we have \( \int_{5}^{1} g(x) \, dx = -8 \).

Step 4: Use Scalar Multiplication Rule for Part (c)

Scalar multiplication of integrals property is \( \int_{a}^{b} c \cdot f(x) \, dx = c \int_{a}^{b} f(x) \, dx \), where \( c \) is a constant. Therefore, \( \int_{1}^{2} 3f(x) \, dx = 3 \times \int_{1}^{2} f(x) \, dx = 3 \times (-4) = -12 \).

Step 5: Apply Additivity Property on Intervals for Part (d)

Use \( \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx = \int_{a}^{b} f(x) \, dx \). We know \( \int_{1}^{5} f(x) \, dx = 6 \) and \( \int_{1}^{2} f(x) \, dx = -4 \). Hence, \( \int_{2}^{5} f(x) \, dx = 6 - (-4) = 6 + 4 = 10 \).

Step 6: Use Linearity Property for Part (e)

By linearity, \( \int_{1}^{5} [f(x) - g(x)] \, dx = \int_{1}^{5} f(x) \, dx - \int_{1}^{5} g(x) \, dx = 6 - 8 = -2 \).

Step 7: Combine Linearity and Scalar Multiplication for Part (f)

Using \( \int_{1}^{5} [c \cdot f(x) - g(x)] \, dx = c \int_{1}^{5} f(x) \, dx - \int_{1}^{5} g(x) \, dx \), where \( c = 4 \), \( \int_{1}^{5}[4f(x) - g(x)] \, dx = 4 \times 6 - 8 = 24 - 8 = 16 \).

Key Concepts

Properties of Integrals

Understanding the properties of definite integrals is crucial when working on calculus problems. Some fundamental properties include:

• Identity Property: The integral of a function over an interval where the upper and lower bounds are the same is zero, i.e., \( \int_{a}^{a} f(x) \, dx = 0 \).
• Reversal of Limits Property: Changing the order of limits changes the sign of the integral, i.e., \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \).

These properties help simplify integrals and evaluate them without direct computation. They make it easy to solve problems just by analyzing the given conditions, allowing us to determine values quickly and efficiently.

Additivity of Integrals

The additivity property of integrals states that you can split an integral over an interval into parts. This is expressed mathematically as:\[\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx\]This property is extremely useful when dealing with complex integral problems where the interval of integration needs to be divided into manageable segments. In practice, when you're given separate integrals over different parts of an interval, you can combine them into a single integral, or vice versa, to find the value of unknown integrals. For instance, if you know:

• \( \int_{1}^{5} f(x) \, dx = 6 \)
• \( \int_{1}^{2} f(x) \, dx = -4 \)

You can find \( \int_{2}^{5} f(x) \, dx \) using additivity: \( 6 - (-4) = 10 \). This method helps in breaking down challenging integrals into simpler parts.

Scalar Multiplication of Integrals

The property of scalar multiplication in integrals is a straightforward yet powerful tool for solving integral problems. The rule is defined as follows:\[\int_{a}^{b} c \cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx\]Where \( c \) is a constant. Essentially, you can pull a scalar factor out of an integral, perform the multiplication after evaluating the integral, and this method vastly simplifies calculations, especially when dealing with expressions involving constants.
For example, if you want to evaluate \( \int_{1}^{2} 3f(x) \, dx \) and you know \( \int_{1}^{2} f(x) \, dx = -4 \), then simply multiply the integral by the scalar: \( 3 \times (-4) = -12 \). This shortcut reduces computational effort while keeping the solution accurate.

Linearity of Integrals

Linearity is one of the most useful properties of integrals. It includes both the additivity of integrals and scalar multiplication and is formulated as:\[\int_{a}^{b} [c_1 \cdot f(x) + c_2 \cdot g(x)] \, dx = c_1 \int_{a}^{b} f(x) \, dx + c_2 \int_{a}^{b} g(x) \, dx\]Here, \( c_1 \) and \( c_2 \) are constants, and it shows that you can distribute the integral across sums or differences of functions. This property helps solve integrals that are linear combinations of known integrals, making complex problems manageable.
Using linearity helps solve problems like \( \int_{1}^{5} [f(x) - g(x)] \, dx \) by separately integrating \( f(x) \) and \( g(x) \), then subtracting the results: \( 6 - 8 = -2 \). Combining linearity with scalar multiplication of integrals offers powerful techniques to efficiently tackle seemingly complex integration problems.