Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove the formula $$ \int\left[f(x) g^{\prime}(x)+g(x) f^{\prime}(x)\right] d x=f(x) g(x)+C $$

Step-by-Step Solution

Answer

The identity is proven using integration by parts, confirming \( \int [f(x) g'(x) + g(x) f'(x)] dx = f(x) g(x) + C \).

Step 1: Understand the Integration by Parts Formula

The formula is a specific case of integration by parts: \( \int u dv = uv - \int v du \). Here, \( f(x) \) and \( g(x) \) are the functions involved. We will use this understanding to prove the formula.

Step 2: Identify Functions for Integration by Parts

The expression \( f(x)g'(x) + g(x)f'(x) \) can be rewritten in the form suitable for integration by parts by letting \( u = f(x) \) and \( dv = g'(x)dx \).

Step 3: Apply Integration by Parts to the First Term

Using \( u = f(x) \) and \( dv = g'(x)dx \), we find \( du = f'(x)dx \) and \( v = g(x) \). The integration by parts formula gives us: \( \int f(x)g'(x) \, dx = f(x)g(x) - \int g(x)f'(x) \, dx \).

Step 4: Recognize the Simplification

Notice that the other part of the original integration problem, \( \int g(x)f'(x) \, dx \), is already provided. Adding \( \int g(x)f'(x) \, dx \) to the result from Step 3 demonstrates the given identity.

Step 5: Combine the Results

Combining the results from the integration by parts and the given integral, we arrive at: \( \int \left[f(x)g'(x) + g(x)f'(x) \right] \, dx = f(x)g(x) \).

Step 6: Add the Integration Constant

Finally, remember every indefinite integral includes a constant of integration \( C \), thus the complete formula is \( \int \left[ f(x)g'(x) + g(x)f'(x) \right] \, dx = f(x)g(x) + C \).

Key Concepts

Definite Integrals

In calculus, a definite integral refers to the evaluation of the integral over a specific interval This process allows us to find the area under a curve between two points on the x-axis. Whenever you are given a definite integral, like \[ \int_a^b f(x) \, dx \], it means you need to calculate the integral using the limits provided, from \(a\) to \(b\). The calculation involves finding the antiderivative of the function first, then applying the Fundamental Theorem of Calculus. This theorem states that:

• If \( F(x) \) is the antiderivative of \( f(x) \), then:
• \[ \int_a^b f(x) \, dx = F(b) - F(a) \]

In practice, definite integrals are crucial for calculating quantities like area, volume, and other cumulative measures where boundaries are defined. Definite integrals also play an essential role in solving real-world problems involving rates of change and accumulation.

Indefinite Integrals

Indefinite integrals are the general form of integration, representing the entirety of antiderivatives of a function. Unlike definite integrals, they do not require boundaries. The expression for indefinite integrals looks like this: \[ \int f(x) \, dx = F(x) + C \], where \( F(x) \) is the antiderivative of \( f(x) \) and \( C \) is the constant of integration.
Indefinite integrals are essentially the reverse operation of differentiation. Antiderivatives are crucial because they help describe the general form of the original function before it was differentiated. The constant \( C \) accounts for the fact that differentiation eliminates constant terms, hence any antiderivative needs an additional constant to fully express the family of solutions for the original function.When working with integration, remember that finding the indefinite integral of a function essentially means finding a function whose derivative gives you the original function. This is why it's sometimes referenced as an "inverse" process of differentiation.

Calculus Techniques

Calculus offers a broad set of techniques for solving problems related to rates of change and the accumulation of quantities. It's split primarily into differential calculus and integral calculus. Differential calculus involves understanding and finding the rate at which things change, generally focused on derivatives and applications such as slopes of curves. On the other hand, integral calculus focuses on the problem of accumulation, dealing with integrals. Integration techniques include:

• Substitution: Used to simplify the integration of composite functions.
• Integration by Parts: Useful for integrating products of functions using a specific formula.
• Partial Fractions: Involves breaking a complex rational function into simpler parts to integrate.

These techniques are valuable tools in calculus because they provide methods to tackle a wide range of integral problems, each with its own unique function types and complexities. Understanding when and how to apply these techniques is key to mastering calculus.

Functions Differentiation

Differentiation is a central concept in calculus that deals with finding the derivative of a function. A derivative represents the rate of change of a function with respect to a variable and is key to understanding patterns of change and behavior of functions.
The process involves applying specific rules such as the power rule, product rule, quotient rule, and chain rule.

• Power Rule: Differentiate functions in the form of \( x^n \).
• Product Rule: Useful when differentiating products of two functions.
• Quotient Rule: Applies when differentiating quotients of two functions.
• Chain Rule: Allows for the differentiation of composite functions.

Differentiation doesn't just tell us about the slope of a curve; it can also give critical insights into function behavior such as inflection points, local minima, and maxima.
By employing differentiation, one can solve a myriad of real-world problems ranging from motion and growth rates to finding optimal solutions in various fields like economics and engineering.Understanding differentiation lays a strong foundation for other calculus concepts, including integration, as we often use derivatives in conjunction with other techniques to analyze functions deeply.