Chapter 6: Problem 20
In Exercises \(6-21,\) find an antiderivative. $$f(t)=\frac{t^{2}+1}{t}$$
Short Answer
Expert verified
The antiderivative of \( f(t) = \frac{t^2 + 1}{t} \) is \( \frac{t^2}{2} + \ln |t| + C \).
Step by step solution
01
Simplify the Function
The given function is \( f(t) = \frac{t^2 + 1}{t} \). To simplify, split the expression into separate terms by dividing each term in the numerator by \( t \). This gives \( f(t) = t + \frac{1}{t} \).
02
Identify Each Term for Integration
Recognize that \( f(t) = t + \frac{1}{t} \) consists of two separate terms: \( t \) and \( \frac{1}{t} \). We will find the antiderivative of each term individually.
03
Find the Antiderivative of \( t \)
The antiderivative of \( t \) is found using the power rule for integration. Increase the exponent by 1 and divide by the new exponent: \( \int t \, dt = \frac{t^{1+1}}{1+1} = \frac{t^2}{2} \).
04
Find the Antiderivative of \( \frac{1}{t} \)
The antiderivative of \( \frac{1}{t} \) is a known result. It is \( \ln |t| \), since the integral \( \int t^{-1} \, dt = \ln |t| \).
05
Combine the Antiderivatives and Include Constant of Integration
Combine the antiderivatives from Steps 3 and 4. Thus, the antiderivative of \( f(t) = t + \frac{1}{t} \) is \( \frac{t^2}{2} + \ln |t| + C \), where \( C \) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration
Integration is a fundamental concept in calculus that helps us find the area under a curve. It is often referred to as the "antiderivative" because it involves reversing the process of differentiation. When confronted with a function such as \( f(t) = \frac{t^{2}+1}{t} \), our goal is to "integrate" it, meaning we want to determine a function whose derivative gives us the original function.
Think of integration as the opposite of differentiation:
Think of integration as the opposite of differentiation:
- Differentiation gives us the rate at which a function changes.
- Integration helps us accumulate the total area or value from a rate.
Power Rule
The power rule for integration is a handy tool that simplifies the process of finding antiderivatives for polynomial terms. It states that to integrate a term of the form \( t^n \), you increase the exponent by one and then divide by the new exponent:
\[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \]However, this rule only applies when \( n eq -1 \). This is because when \( n = -1 \), the term becomes \( \frac{1}{t} \), which requires a different approach.
In our example, when faced with \( t \), we recognize it as \( t^1 \). Applying the power rule gives us the antiderivative:
\[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \]However, this rule only applies when \( n eq -1 \). This is because when \( n = -1 \), the term becomes \( \frac{1}{t} \), which requires a different approach.
In our example, when faced with \( t \), we recognize it as \( t^1 \). Applying the power rule gives us the antiderivative:
- \( n = 1 \)
- New exponent: \( 1 + 1 = 2 \)
- Antiderivative: \( \frac{t^2}{2} \)
Natural Logarithm
When an integration task involves \( \frac{1}{t} \), we encounter a special case that the power rule cannot handle directly. This is where the natural logarithm comes into play. The term \( \frac{1}{t} \) is equivalent to \( t^{-1} \), and its antiderivative is the natural logarithm of the absolute value of \( t \):
\[ \int \frac{1}{t} \, dt = \ln |t| + C \]The beauty of using the natural logarithm in integration is that it allows us to handle terms that involve division by a variable, which are common in calculus problems.
In our specific problem, recognizing \( \frac{1}{t} \) helps us apply the logarithmic rule directly, ensuring we integrate the term correctly. Remember, the absolute value in \( \ln |t| \) is essential to account for both positive and negative values of \( t \).
\[ \int \frac{1}{t} \, dt = \ln |t| + C \]The beauty of using the natural logarithm in integration is that it allows us to handle terms that involve division by a variable, which are common in calculus problems.
In our specific problem, recognizing \( \frac{1}{t} \) helps us apply the logarithmic rule directly, ensuring we integrate the term correctly. Remember, the absolute value in \( \ln |t| \) is essential to account for both positive and negative values of \( t \).
Constant of Integration
In the process of finding antiderivatives or integrals, we always append a constant of integration, usually denoted by \( C \). This step is crucial because integration represents a family of functions, not just a single function.
When we differentiate a constant, we get zero, meaning that any constant term could have been part of the original function prior to differentiation – this is why the constant is an integral component of the solution.
\[ \frac{t^2}{2} + \ln |t| + C \]This expression represents the general solution of the integration problem, showcasing the elegance and completeness of calculus in describing curves and their properties.
When we differentiate a constant, we get zero, meaning that any constant term could have been part of the original function prior to differentiation – this is why the constant is an integral component of the solution.
- It accounts for the "lost" constant during differentiation.
- Allows integration solutions to encompass all possible original functions.
\[ \frac{t^2}{2} + \ln |t| + C \]This expression represents the general solution of the integration problem, showcasing the elegance and completeness of calculus in describing curves and their properties.