Problem 35 - Calculus

Question

If \(d f / d x=v(x)\) then \(\int v(x-1) d x=\ldots\) and \(\int v(x / 2) d x=\)

Step-by-Step Solution

Answer

\( \int v(x-1) \, dx = f(x-1) + C \) and \( \int v(x/2) \, dx = 2f(x/2) + C \).

Step 1: Understanding the problem

We are given that \( \frac{df}{dx} = v(x) \), which means that \( v(x) \) is the derivative of a function \( f(x) \). Our task is to find \( \int v(x-1) \, dx \) and \( \int v(x/2) \, dx \).

Step 2: Substitution for the first integral

To find \( \int v(x-1) \, dx \), we make a substitution. Let \( u = x - 1 \). Then, \( du = dx \). The integral becomes \( \int v(u) \, du \). Since \( v(u) = \frac{df}{du} \), the integral simplifies to \( f(u) + C \), where \( C \) is the constant of integration.

Step 3: Back-substitution for the first integral

Back-substitute \( u = x - 1 \) into the result to return to the original variable: \( f(u) = f(x-1) \). Therefore, \( \int v(x-1) \, dx = f(x-1) + C \).

Step 4: Substitution for the second integral

To find \( \int v(x/2) \, dx \), we perform another substitution. Let \( u = x/2 \), then \( x = 2u \) and \( dx = 2 \, du \). The integral becomes \( \int v(u) \, 2 \, du \).

Step 5: Solve the substituted second integral

The integral \( \int 2v(u) \, du \) relates to the function \( f(u) \), given \( v(u) = \frac{df}{du} \), it simplifies to \( 2f(u) + C \).

Step 6: Back-substitution for the second integral

Substitute back \( u = x/2 \): We get \( f(u) = f(x/2) \). Therefore, \( \int v(x/2) \, dx = 2f(x/2) + C \).

Key Concepts

Definite Integrals

A definite integral is a way to calculate the area under a curve between two specified points. In mathematical terms, if you have a function \( f(x) \), the definite integral from \( a \) to \( b \) is written as \( \int_{a}^{b} f(x) \, dx \). This type of integral gives a numerical value that represents the accumulated sum of areas between the function and the x-axis over the interval \([a, b]\).

• Key Properties: The bounds \( a \) and \( b \) indicate the exact range for calculation.
• Fundamental Theorem of Calculus: Establishes a relationship between differentiation and integration. It states that if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).

Understanding definite integrals is crucial for determining areas, solving real-life problems involving rates of change, and many other applied mathematics scenarios. An important aspect of definite integrals is that the same area can be calculated using different techniques, including substitution, as we explore integration by substitution techniques that involve changing variables.

Indefinite Integrals

An indefinite integral, on the other hand, does not have limits of integration and generally represents a family of functions. It includes all the antiderivatives of a given function. Mathematically, it is expressed as \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is an antiderivative and \( C \) is the constant of integration.

• Antiderivative: The inverse process of differentiation. It finds a function \( F(x) \) whose derivative is \( f(x) \).
• Constant of Integration: Since differentiation of a constant is zero, \( C \) accounts for any constant that was lost during differentiation.

In the original exercise, solving \( \int v(x-1) \, dx \) and \( \int v(x/2) \, dx \) are classic examples of finding indefinite integrals:

• Example 1: For \( \int v(x-1) \, dx \), through substitution, we simplify it to \( f(x-1) + C \).
• Example 2: Similarly, for \( \int v(x/2) \, dx \), substitution leads to \( 2f(x/2) + C \).

Indefinite integrals are crucial for developing formulas for physics and engineering problems, optimizing functions, and understanding rates of change in calculus.

Change of Variables

The technique of change of variables, often employed in integration as substitution, simplifies complex integrals by transforming them into simpler forms. The principle works by introducing a new variable to replace an expression in the integrand, making the resulting integral easier to solve.

• Substitution: This means replacing a part of the integral, such as \( x-1 \) or \( x/2 \), with a single variable, \( u \).
• Integration Process: This involves finding \( du \) and expressing \( dx \) in terms of \( du \), simplifying the integrand.

In the exercise example, we used:

• Case 1: Set \( u = x - 1 \), which simplified \( \int v(x-1) \, dx \) to \( \int v(u) \, du \).
• Case 2: Set \( u = x/2 \), turning \( \int v(x/2) \, dx \) into \( \int 2v(u) \, du \).

Substitutions help bridge complex functions to simpler forms, which are easier to integrate. This method is not just limited to indefinite integrals but is also crucial for definite integrals. Understanding when and how to effectively use substitution is key to mastering advanced calculus and tackling a variety of integrative problems.