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Find the indefinite (or definite) integral. $$ \int_{0}^{1} \frac{d t}{3+2 t} $$

Short Answer

Expert verified
\( \frac{1}{2} \ln \left( \frac{5}{3} \right) \)

Step by step solution

01

Identify the Integral Type

We observe that the given integral is a definite integral with the limits from 0 to 1. The integrand is \( \frac{1}{3+2t} \), which suggests that we might use a substitution or recognition method.
02

Use Substitution Method

To simplify the integral, let's use substitution. Set \( u = 3 + 2t \). Then, differentiate \( u \) with respect to \( t \): \( \frac{du}{dt} = 2 \). This implies \( dt = \frac{du}{2} \). Also, change the limits of integration: when \( t = 0, u = 3 \) and when \( t = 1, u = 5 \).
03

Change the Variable in the Integral

Replace \( t \) in the integral with \( u \). The integral becomes: \[ \int_{3}^{5} \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int_{3}^{5} \frac{du}{u} \]
04

Integrate the Expression

Integrate the expression \( \int \frac{du}{u} \). The antiderivative is \( \ln|u| + C \). Thus, \[ \frac{1}{2} \int_{3}^{5} \frac{du}{u} = \frac{1}{2} \left[ \ln|u| \right]_{3}^{5} \]
05

Evaluate the Definite Integral

Now, evaluate the expression from Step 4 using the limits 3 and 5:\[ \frac{1}{2} \left( \ln|5| - \ln|3| \right) = \frac{1}{2} \left( \ln{5} - \ln{3} \right) \]
06

Simplify the Expression

Use the property of logarithms, \( \ln{a} - \ln{b} = \ln{\frac{a}{b}} \), to simplify the expression:\[ \frac{1}{2} \ln{\frac{5}{3}} \]
07

Final Answer

The value of the integral \( \int_{0}^{1} \frac{dt}{3+2t} \) is:\[ \frac{1}{2} \ln \left( \frac{5}{3} \right) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The Substitution Method is a technique in calculus used to simplify the integration process by changing the variable of integration to a new variable. This method is particularly useful when dealing with composite functions, where direct integration seems complex or not straightforward. In our case, the substitution of a new variable, say \( u \), replaces a part of the function to simplify the integral.
Let's consider the given integral \( \int_{0}^{1} \frac{d t}{3+2 t} \). Here, the term \( 3+2t \) can be replaced by \( u \). We then differentiate \( u \) with respect to \( t \), obtaining \( \frac{du}{dt} = 2 \). This implies \( dt = \frac{du}{2} \). The original bounds of integration, \( t = 0 \) and \( t = 1 \), create corresponding bounds for \( u \), changing them to \( u = 3 \) and \( u = 5 \) respectively.
This transformation simplifies the integral to \( \frac{1}{2} \int_{3}^{5} \frac{du}{u} \), a much easier form to integrate.
Antiderivative
An antiderivative of a function is another function whose derivative is the original function. In integration, finding the antiderivative is key to solving definite or indefinite integrals. It involves determining a function that, when differentiated, returns the integrand.
For our exercise, once the substitution method simplifies the integral to \( \int \frac{du}{u} \), finding the antiderivative becomes straightforward. The antiderivative of \( \frac{1}{u} \) is \( \ln|u| + C \), where \( C \) is the constant of integration. Since we are dealing with a definite integral, the integration constant \( C \) is not needed in the calculation.
Therefore, the integral expression \( \frac{1}{2} \left[ \ln|u| \right]_{3}^{5} \) evaluates the natural logarithm of \( u \) at the limits \( 3 \) and \( 5 \).
Logarithm Property
In the context of integration, the properties of logarithms often help to simplify results and calculations. One critical property is that of difference in the logarithms: \( \ln{a} - \ln{b} = \ln\left(\frac{a}{b}\right) \).
By understanding and using this property, we reduce expressions like \( \ln|5| - \ln|3| \) to a single logarithm term \( \ln\left(\frac{5}{3}\right) \). Simplifying expressions in this way not only makes them easier to interpret, but it also aligns with common practices in mathematical problem-solving, keeping results concise.
So, upon evaluating the integral expression, applying this property results in the final form: \( \frac{1}{2} \ln\left(\frac{5}{3}\right) \), which is a neater representation.
Integration Limits
Integration limits are crucial when dealing with definite integrals. They determine the bounds over which the function is integrated, providing a specific numerical result rather than a general formula with a constant.
As part of the substitution process, changing integration limits is necessary to reflect the transformation of variables. Initially, the limits for \( t \) were \( 0 \) and \( 1 \). After substituting \( u = 3 + 2t \), the limits became \( 3 \) when \( t = 0 \), and \( 5 \) when \( t = 1 \).
It's important to remember: after a substitution, always adjust limits for the new variable to accurately calculate the definite integral, ensuring the solution aligns with the transformed integral.

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