Problem 67 - Applied Calculus

Question

Each equation follows from the integration by parts formula by replacing \(u\) by \(f(x)\) and \(v\) by a particular function. What is the function \(v ?\) \(\int f(x) e^{x} d x=f(x) e^{x}-\int e^{x} f^{\prime}(x) d x\)

Step-by-Step Solution

Answer

The function \(v\) is \( e^{x} \).

Step 1: Identify Integration by Parts Formula

The integration by parts formula is expressed as \( \int u \cdot dv = uv - \int v \cdot du \). We need to match this formula with the given equation \( \int f(x) e^{x} \, dx = f(x) e^{x} - \int e^{x} f^{\prime}(x) \, dx \).

Step 2: Assign \(u\) and \(dv\)

In the given equation, \(u = f(x)\) and consequently \(du = f^{\prime}(x) \, dx\). The term \(dv\) in the integration by parts would correspond to the other multiplicative part of the integrand, which is \( e^{x} \, dx \).

Step 3: Find \(v\) from \(dv\)

Since \(dv = e^{x} \, dx\), integrate this to find \(v\). The integration of \( e^{x} \, dx \) is \( v = e^{x} \).

Step 4: Verify the Integration by Parts

Plug \( u = f(x) \), \( du = f^{\prime}(x) \, dx \), \( v = e^{x} \), and \( dv = e^{x} \, dx \) back into the formula \( \int u \cdot dv = uv - \int v \cdot du \). This simplifies to \( \int f(x) \, e^{x} \, dx = f(x) \, e^{x} - \int e^{x} \, f^{\prime}(x) \, dx \), which matches the given equation.

Key Concepts

Integration

Integration is a fundamental concept in calculus, which revolves around the idea of finding the area under a curve. When we integrate a function, we try to reverse the process of differentiation, essentially summing up an infinite number of infinitesimally small quantities to find a whole.

To understand integration, imagine you are stacking very thin slices of bread, where each slice represents a small portion of the area under a curve. The whole loaf represents the total area. This is similar to how integration sums up tiny parts to get the complete area.

• **Definition**: Integration is the process of calculating the integral of a function, usually represented as \( \int f(x) \, dx \).
• **Application**: It is frequently used to find areas, volumes, central points, and many other useful things.

There are different techniques for integration, such as substitution, integration by parts, and partial fraction decomposition. Integration by parts, the technique used in the exercise, is particularly useful when dealing with the product of functions. This method transforms the integral of a product of functions into simpler forms that we already know how to integrate. Understanding the nature and properties of these techniques helps greatly in solving complex integral problems.

Integrals

Integrals can be thought of as the quantities or results we obtain from the integration process. The concept of an integral helps us deal with whole quantities that are accumulations of infinitely small contributions. Integrals are split into definite and indefinite integrals.

• **Indefinite Integrals**: Represented as \( \int f(x) \, dx \), they result in a family of functions \( F(x) + C \), where \( C \) is the constant of integration. This form does not specify limits and thus represents an infinite set of solutions.
• **Definite Integrals**: Have specific limits from \( a \) to \( b \) and are represented as \( \int_{a}^{b} f(x) \, dx \). These compute the exact area under the curve between two points and are evaluated to a number.

In the context of the exercise mentioned, the definite or indefinite nature of the integral isn't specified. However, the primary focus is on the integration by parts technique, where you identify functions \( u \) and \( dv \) from the integrand to simplify the computation. It's crucial to distinguish between the applicability of definite and indefinite integrals to solve real-world problems.

Calculus

Calculus is the branch of mathematics focused on change and motion. It comprises two main concepts: differentiation and integration. These two operations are fundamental and inverse processes that describe how functions behave and how they can be analyzed.

• **Differentiation**: Concerned with finding the rate at which things change, typically leading to the concept of derivatives. It's the mathematical way of understanding how something moves or grows instantly.
• **Integration**: As mentioned before, is about accumulating quantities, often used to determine areas, volumes, and other values where aggregation or summation is involved.

Integration by parts is a rule stemming from integration that is useful when integrating the product of two functions. In the exercise you learned, this rule helped transform a complex integral problem into a more manageable form by breaking down the integrand component.
Understanding the seamless relationship between differentiation and integration is key in calculus. These processes allow the conversion of real-world situations into mathematical equations, which can then be analyzed to make predictions and solve problems.