Problem 14
Question
Find the integrals in problems. Check your answers by differentiation. $$ \int \frac{d y}{y+5} $$
Step-by-Step Solution
Verified Answer
The integral is \( \ln |y+5| + C \).
1Step 1: Identify the Indefinite Integral
The goal is to find the integral \( \int \frac{d y}{y+5} \). This requires recognizing that it falls under the category of integral formulas for logarithmic functions.
2Step 2: Apply the Integration Formula
Since the integral \( \int \frac{1}{u} \, du \) equals \( \ln |u| + C \), where \( C \) is the constant of integration, we can write: \( \int \frac{d y}{y+5} = \ln |y+5| + C \).
3Step 3: Check the Result by Differentiation
To verify the solution, differentiate \( \ln |y+5| + C \) with respect to \( y \). The derivative \( \frac{d}{dy}[ \ln |y+5| + C ] = \frac{1}{y+5} \), which confirms the original integrand.
Key Concepts
Indefinite IntegralLogarithmic FunctionsDifferentiation
Indefinite Integral
Indefinite integrals, often represented as \( \int f(x) \, dx \), are foundational in calculus. They signify a family of functions whose derivative leads back to \( f(x) \). Unlike definite integrals that result in numerical values over a specific interval, indefinite integrals include a constant of integration, denoted as \( C \). This constant represents the unknown initial value since differentiation of any constant yields zero.
When dealing with indefinite integrals, one can express the solution as a general formula. For example, when you integrate \( \frac{1}{x} \), the result is \( \ln |x| + C \), derived from the integral formula \( \int \frac{1}{x} \, dx = \ln |x| + C \). This technique applies to many functions, turning complex expressions into manageable terms by reversing differentiation processes.
Mastering indefinite integrals involves recognizing patterns and understanding fundamental integration rules to apply them appropriately. These skills enable solving various mathematical problems, modeling real-world phenomena, and simplifying complex expressions.
When dealing with indefinite integrals, one can express the solution as a general formula. For example, when you integrate \( \frac{1}{x} \), the result is \( \ln |x| + C \), derived from the integral formula \( \int \frac{1}{x} \, dx = \ln |x| + C \). This technique applies to many functions, turning complex expressions into manageable terms by reversing differentiation processes.
Mastering indefinite integrals involves recognizing patterns and understanding fundamental integration rules to apply them appropriately. These skills enable solving various mathematical problems, modeling real-world phenomena, and simplifying complex expressions.
Logarithmic Functions
Logarithmic functions play a crucial role in calculus, particularly when integrating functions like \( \int \frac{1}{u} \, du \). These functions serve as the inverse of exponential functions and are critical in processes involving growth and decay. The standard form is \( \log_b(x) \), where \( b \) is the base of the logarithm. However, the most commonly used form in calculus is the natural logarithm, \( \ln(x) \), which uses base \( e \).In the context of integration, logarithmic functions help in cases involving division by linear expressions. The formula \( \int \frac{1}{u} \, du = \ln |u| + C \) simplifies integrals of functions like \( \frac{1}{x} \) or \( \frac{1}{y + 5} \).
This formula allows us to handle expressions that might otherwise appear challenging. For example, integrating \( \frac{1}{y+5} \) results in \( \ln |y+5| + C \), directly applying the known logarithmic integral property. Recognizing scenarios where logarithmic functions are applicable helps in solving integrals efficiently and ensures a thorough understanding of continuous growth and decay phenomena.
This formula allows us to handle expressions that might otherwise appear challenging. For example, integrating \( \frac{1}{y+5} \) results in \( \ln |y+5| + C \), directly applying the known logarithmic integral property. Recognizing scenarios where logarithmic functions are applicable helps in solving integrals efficiently and ensures a thorough understanding of continuous growth and decay phenomena.
Differentiation
Differentiation, the process of finding a derivative, is the cornerstone of calculus. It measures how a function changes as its input changes. When you differentiate \( f(x) \), the result, \( f'(x) \), is a function representing the rate of change or the slope at any given point on the curve of \( f(x) \).In verifying integration results, differentiation is indispensable. To check the integration \( \int \frac{d y}{y+5} = \ln |y+5| + C \), you simply differentiate \( \ln |y+5| + C \). The derivative of \( \ln |y+5| \) is \( \frac{1}{y+5} \), because the derivative of \( \ln |u| \) is \( \frac{1}{u} \), and \( C \) as a constant vanishes \( (\frac{d}{dy}[C] = 0) \). This confirms that \( \ln |y+5| + C \) is indeed the antiderivative of \( \frac{1}{y+5} \).
Through differentiation, we can test our integration results, offering a straightforward approach to ensure correctness. This dual relationship between integration and differentiation forms a robust framework, enabling deeper exploration and understanding of mathematical and physical systems.
Through differentiation, we can test our integration results, offering a straightforward approach to ensure correctness. This dual relationship between integration and differentiation forms a robust framework, enabling deeper exploration and understanding of mathematical and physical systems.
Other exercises in this chapter
Problem 14
Find the integrals. $$ \int_{3}^{5} x \cos x d x $$
View solution Problem 14
Using the Fundamental Theorem, evaluate the definite integrals in problem exactly. $$ \int_{2}^{5}\left(x^{3}-\pi x^{2}\right) d x $$
View solution Problem 14
Find an antiderivative. $$ q(z)=\sqrt{z} $$
View solution Problem 15
Use integration by parts twice to evaluate the integral. $$ \int t^{2} e^{5 t} d t $$
View solution