Question

For the following exercises, find the definite or indefinite integral. $$ \int_{0}^{1} \frac{d t}{3+2 t} $$

Short answer

The integral evaluates to \( \frac{1}{2} \ln \frac{5}{3} \).

Step-by-step solution

Identify the Integral Type

We need to find the definite integral \( \int_{0}^{1} \frac{d t}{3+2t} \). This is a definite integral because it has specified limits (0 to 1).

Simplify the Integrand

The integrand is \( \frac{1}{3+2t} \). This is a simple rational function which can be integrated using substitution.

Use Substitution

Let \( u = 3 + 2t \). Then, \( \frac{du}{dt} = 2 \), so \( dt = \frac{du}{2} \). Also, when \( t = 0, u = 3 \) and when \( t = 1, u = 5 \).

Rewrite the Integral

Substituting in terms of \( u \), the integral becomes \( \int_{3}^{5} \frac{1}{u} \cdot \frac{du}{2} \). This simplifies to \( \frac{1}{2} \int_{3}^{5} \frac{1}{u} \, du \).

Integrate

The integral \( \int \frac{1}{u} \, du \) is \( \ln|u| + C \). So, \( \frac{1}{2} \left[ \ln |u| \right]_{3}^{5} \) is evaluated.

Evaluate the Definite Integral

Calculate \[ \frac{1}{2} \left( \ln|5| - \ln|3| \right) = \frac{1}{2} (\ln 5 - \ln 3) \]. This simplifies to \( \frac{1}{2} \ln \frac{5}{3} \).

Final Answer

The definite integral from 0 to 1 of \( \frac{1}{3+2t} \ dt \) is \( \frac{1}{2} \ln \frac{5}{3} \).

Key concepts

Substitution Method

The substitution method is a powerful technique used to simplify the process of integration. It involves substituting a part of the integrand with a new variable, typically using a simple algebraic relationship. This makes the integration easier to handle. For example, in the exercise provided, we need to integrate the function \( \int_{0}^{1} \frac{d t}{3+2t} \). Here, the term \( 3 + 2t \) is inside the denominator and is a good candidate for substitution.

Here's how it works in steps:

• Choose a substitution: Set \( u = 3 + 2t \).
• Differentiate to find \( du \): We differentiate to get \( \frac{du}{dt} = 2 \), which means \( dt = \frac{du}{2} \).
• Change the limits of the integral: When \( t = 0 \), \( u = 3 \). When \( t = 1 \), \( u = 5 \).
• Substitute in the integral: Replace \( t \) with \( u \) and express \( dt \) in terms of \( du \), so the integral becomes \( \int_{3}^{5} \frac{1}{u} \cdot \frac{du}{2} \).

Substitution simplifies the integrand, making the function easier to integrate. This method is particularly helpful for beginners because it can transform complex-looking integrals into a more familiar form.

Rational Function Integration

Rational functions are ratios of polynomials. They are common in calculus and often appear in integrals. Integrating rational functions can sometimes be straightforward, especially if the denominator can be easily manipulated or decomposed using algebraic techniques. In this exercise, the function \( \frac{1}{3+2t} \) is a simple rational function.

For rational function integration, we look for ways to reduce or simplify the function:

• Direct substitution like we used in this problem is often effective.
• Partial fraction decomposition is another method, useful for more complex rational functions.

In the provided exercise, the function was managed elegantly through substitution by simplifying the expression to \( \frac{1}{u} \), which is straightforward to integrate. The task revolves around finding suitable substitutions or transformations that make the problem solvable by familiar integration rules.

Natural Logarithm Integration

Natural logarithm integration is a fundamental skill in calculus, often appearing when integrating functions of the form \( \frac{1}{x} \). This form leads directly to the natural logarithm function \( \ln|x| \), which is the antiderivative. In this case, after substitution, our integral becomes \( \frac{1}{2} \int_{3}^{5} \frac{1}{u} \, du \).

Here's what happens:

• The integral of \( \frac{1}{u} \) is \( \ln|u| \). This is a standard result widely used in calculus.
• Since the definite integral is evaluated between the limits \( 3 \) and \( 5 \), we apply the Fundamental Theorem of Calculus.

We evaluate \( \frac{1}{2} (\ln 5 - \ln 3) \). Also, properties of logarithms such as \( \ln a - \ln b = \ln \frac{a}{b} \) simplify the expression to \( \frac{1}{2} \ln \frac{5}{3} \).

Understanding the link between the integrals of basic rational functions and natural logs is crucial. This makes tackling more complex integration problems much more manageable, leveraging known derivatives and properties of logarithms.