Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove the formula $$ \int\left[f(x) g^{\prime}(x)+g(x) f^{\prime}(x)\right] d x=f(x) g(x)+C $$ Hint: See the box in the margin next to Theorem A.

Step-by-Step Solution

Answer

By the product rule, the integral is \( f(x)g(x) + C \).

Step 1: Understand the Question

The task is to prove that the integral \( \int \left[ f(x) g'(x) + g(x) f'(x) \right] \, dx \) is equal to \( f(x)g(x) + C \), where \( C \) is the constant of integration.

Step 2: Recall the Product Rule from Calculus

The product rule states that the derivative of a product \( f(x)g(x) \) is \( f'(x)g(x) + f(x)g'(x) \). This is a key concept that relates to the problem.

Step 3: Rewrite the Expression

Notice that \( f(x) g'(x) + g(x) f'(x) \) can be viewed as the derivative of the product \( f(x) g(x) \) using the product rule.

Step 4: Set Up the Integral

Now consider integrating the derivative of a function, i.e., \( \int \left( \frac{d}{dx}[f(x) g(x)] \right) \, dx \). According to the Fundamental Theorem of Calculus, this is simply \( f(x)g(x) + C \).

Step 5: Conclude the Solution

Since integrating the expression using the product rule directly results in \( f(x)g(x) + C \), it follows that the original integral \( \int \left[ f(x) g'(x) + g(x) f'(x) \right] \, dx = f(x)g(x) + C \). This proves the statement.

Key Concepts

calculus

Calculus is a branch of mathematics that deals with the study of change. There are two primary branches in calculus: differential calculus and integral calculus. Differential calculus focuses on finding the rate of change of a function, while integral calculus is concerned with the accumulation of quantities. In the context of this exercise, calculus helps in forming integrals and differentiating functions, using concepts like the product rule and integration by parts. The problem at hand uses these foundational ideas to prove an expression involving derivatives and integrals. Calculus is not only essential for theoretical mathematics, but also has applications in physics, engineering, economics, and beyond.

fundamental theorem of calculus

The Fundamental Theorem of Calculus (FTC) is a central concept that connects differentiation and integration, two of the main operations in calculus. It consists of two parts:

• The first part states that if a function is continuous over an interval, then the integral over that interval can be measured using its antiderivative.
• The second part states that the derivative of an integral function is the original function itself.

In our exercise, the FTC is used to equate the integral of the derivative of a product to the product itself plus a constant, effectively simplifying the problem. The FTC elegantly bridges the gap between the abstract concept of integration and the more tangible idea of derivatives.

product rule

The product rule is a formula used to find the derivative of the product of two functions. It states that if you have two functions, say \( f(x) \) and \( g(x) \), then the derivative of their product is given by:\[(fg)' = f'g + fg'\]In plain terms, if you are differentiating a product, you need to take the derivative of the first function times the second function, plus the first function times the derivative of the second. In this exercise, the product rule helps transform the expression \( f(x) g'(x) + g(x) f'(x) \) into a form that represents the derivative of the function \( f(x)g(x) \). This step is critical for leveraging the Fundamental Theorem of Calculus to solve the integral.

indefinite integral

An indefinite integral represents a family of functions and is symbolized by the integral sign \( \int \), followed by the function to integrate and the differential \( dx \). Unlike a definite integral, it does not have upper and lower limits for integration and includes an arbitrary constant \( C \). This constant accounts for the fact that there are countless antiderivatives for a given function, differing by a constant value. In our exercise, the indefinite integral \( \int \left[ f(x) g'(x) + g(x) f'(x) \right] \, dx \) is evaluated to prove that it equals \( f(x)g(x) + C \). The constant \( C \) is essential because, without limits, we need to reflect any variations in possible solutions.