Problem 9 - Thomas Calculus

Question

Suppose that \(f\) and \(g\) are integrable and that $$\int_{1}^{2} f(x) d x=-4, \int_{1}^{5} f(x) d x=6, \int_{1}^{5} g(x) d x=8$$ Use the rules in Table 5.3 to find $$ \begin{array}{ll}{\text { a. } \int_{2}^{2} g(x) d x} & {\text { b. } \int_{5}^{1} g(x) d x} \\ {\text { c. } \int_{1}^{2} 3 f(x) d x} & {\text { d. } \int_{2}^{5} f(x) d x} \\ {\text { e. } \int_{1}^{5}[f(x)-g(x)] d x} & {\text { f. } \int_{1}^{5}[4 f(x)-g(x)] d x}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Properties of Integrals

First, it's important to remember that the integral of any function from a to a is zero, that swapping the limits of an integral inverts its sign, and that the integral respects linearity (you can pull constants out and break additions into separate integrals). These properties will be used to solve our integrals.

Step 1: Solving for \( \int_{2}^{2} g(x) \, dx \)

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

Step 2: Solving for \( \int_{5}^{1} g(x) \, dx \)

Using the property that swapping the limits of integration changes the sign of the integral:\[ \int_{5}^{1} g(x) \, dx = - \int_{1}^{5} g(x) \, dx, \]we know from the given that \( \int_{1}^{5} g(x) \, dx = 8 \), hence, \( \int_{5}^{1} g(x) \, dx = -8 \).

Step 3: Solving for \( \int_{1}^{2} 3 f(x) \, dx \)

Using the property that constants can be factored out of integrals:\[ \int_{1}^{2} 3f(x) \, dx = 3 \int_{1}^{2} f(x) \, dx. \]We're given \( \int_{1}^{2} f(x) \, dx = -4 \), hence, \( 3 \times (-4) = -12 \). So, \( \int_{1}^{2} 3 f(x) \, dx = -12 \).

Step 4: Solving for \( \int_{2}^{5} f(x) \, dx \)

Using the additive property of integrals:\[ \int_{1}^{5} f(x) \, dx = \int_{1}^{2} f(x) \, dx + \int_{2}^{5} f(x) \, dx. \]Given \( \int_{1}^{5} f(x) \, dx = 6 \) and \( \int_{1}^{2} f(x) \, dx = -4 \), we can solve for \( \int_{2}^{5} f(x) \, dx \) by rearranging:\[ 6 = -4 + \int_{2}^{5} f(x) \, dx \text{ implies } \int_{2}^{5} f(x) \, dx = 6 - (-4) = 10. \]

Step 5: Solving for \( \int_{1}^{5} [f(x) - g(x)] \, dx \)

Using the property \[ \int_{1}^{5} [f(x) - g(x)] \, dx = \int_{1}^{5} f(x) \, dx - \int_{1}^{5} g(x) \, dx \]and given \( \int_{1}^{5} f(x) \, dx = 6 \) and \( \int_{1}^{5} g(x) \, dx = 8 \), we find:\[ \int_{1}^{5} [f(x) - g(x)] \, dx = 6 - 8 = -2. \]

Step 6: Solving for \( \int_{1}^{5} [4f(x) - g(x)] \, dx \)

Apply linearity to get:\[ \int_{1}^{5} [4f(x) - g(x)] \, dx = 4 \int_{1}^{5} f(x) \, dx - \int_{1}^{5} g(x) \, dx. \]Given \( \int_{1}^{5} f(x) \, dx = 6 \) and \( \int_{1}^{5} g(x) = 8 \), solve:\[ 4 \times 6 - 8 = 24 - 8 = 16. \]

Key Concepts

Definite Integrals

Definite integrals are an essential concept in calculus. They are used to calculate the accumulation of quantities, like areas under curves, over a particular interval. The definite integral of a function \(f(x)\) from \(a\) to \(b\) is denoted as \(\int_{a}^{b} f(x) \, dx\). Here, \(a\) and \(b\) are the limits of integration. The process of determining a definite integral results in a numerical value, unlike indefinite integrals, which yield a function.

In practical terms, definite integrals allow us to measure the total accumulation from point \(a\) to point \(b\). This accumulation might represent the total distance traveled, the total volume under a surface, or other similar cumulative quantities. Moreover, the value of a definite integral is affected by the behavior of the function and the specified interval. Understanding these integrals helps us apply calculus concepts in various fields, including physics and engineering.

Properties of Integrals

The powerful properties of integrals make them friendly to use, providing shortcuts to solving complex problems. One important property is that the integral of a function from a point to itself is always zero: \(\int_{a}^{a} f(x) \, dx = 0\). This describes a scenario where the start and end points of the integration are the same, resulting in no area under the curve.

Another fundamental property involves swapping limits of integration, which changes the sign of the integral. If you have \(\int_{a}^{b} f(x) \, dx\), then \(\int_{b}^{a} f(x) \, dx = -\int_{a}^{b} f(x) \, dx\). This is useful when reversing the direction of integration.

• Additive Property: Integrals are additive, meaning \(\int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx = \int_{a}^{b} f(x) \, dx\).
• Linearity Property: Allows for constants to be factored and expressions broken into parts, facilitating simpler integration methods.

These integral properties are essential tools, offering strategies to simplify calculations and solve integrals more efficiently.

Integration Techniques

Integration techniques are strategic methods developed to solve integrals effectively. When facing definite integrals, these techniques make the process manageable, especially when basic methods are not sufficient.

• Substitution Method: This technique involves changing variables to simplify the integration of complex functions. It is especially useful when dealing with composite functions.
• Integration by Parts: A favorite for products of functions, it is derived from the product rule for differentiation.

Each technique involves transforming the original integrand into a form that is more convenient to integrate. They rely heavily on recognizing patterns within the functions, which require practice and familiarity. Mastering these techniques equips students with essential tools to tackle a wide range of integral problems confidently.

Linearity of Integrals

The linearity of integrals is a core principle that greatly simplifies the process of integration. It states that integrals can be distributed over addition and subtraction, and constants can be factored out of the integrals. Mathematically, it is expressed as:

• \(\int_{a}^{b} [c \cdot f(x) + d \cdot g(x)] \, dx = c \int_{a}^{b} f(x) \, dx + d \int_{a}^{b} g(x) \, dx\)

This property allows us to take constants out of the integral and deal with each function separately, thus making integration easier to handle.

For example, given \(\int_{a}^{b} [3f(x) - g(x)] \, dx\), we can split it into \[3\int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx\], thereby simplifying complex functions into more manageable parts. This principle is often used when combining or comparing functions over the same interval, providing a clear path toward solution.