Chapter 5: Problem 35
If \(d f / d x=v(x)\) then \(\int v(x-1) d x=\ldots\) and \(\int v(x / 2) d x=\)
Short Answer
Expert verified
\( \int v(x-1) \, dx = f(x-1) + C \) and \( \int v(x/2) \, dx = 2f(x/2) + C \).
Step by step solution
01
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 \( \int v(x/2) \, dx \).
02
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.
03
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, \( \int v(x-1) \, dx = f(x-1) + C \).
04
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 \( \int v(u) \, 2 \, du \).
05
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 \( 2f(u) + C \).
06
Back-substitution for the second integral
Substitute back \( u = x/2 \): We get \( f(u) = f(x/2) \). Therefore, \( \int v(x/2) \, dx = 2f(x/2) + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
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.
- 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 \( 2f(x/2) + C \).
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.
- 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 \( \int 2v(u) \, du \).