Problem 46 - Applied Calculus

Question

Find the indefinite integrals. $$ \int e^{2 t} d t $$

Step-by-Step Solution

Answer

\( \int e^{2t} \, dt = \frac{1}{2} e^{2t} + C \).

Step 1: Identify the Form of the Integral

The given integral is \( \int e^{2t} \, dt \). This is an exponential function where the exponent is a linear function of \( t \). The derivative of \( e^{kt} \) with respect to \( t \) is \( ke^{kt} \), which helps us in solving the integral by using substitution if needed.

Step 2: Perform Substitution (if needed)

In this case, we need to perform a substitution to simplify the integration. Let \( u = 2t \). Then, the derivative \( du = 2 \, dt \), or \( dt = \frac{1}{2} \, du \). Now substitute these into the integral, which gives us \( \int e^{u} \cdot \frac{1}{2} \, du \).

Step 3: Integrate the Simplified Expression

Now integrate the expression \( \int \frac{1}{2} e^{u} \, du \). The integral of \( e^u \) with respect to \( u \) is \( e^u \). Thus, the expression becomes \( \frac{1}{2} e^u + C \), where \( C \) is the constant of integration.

Step 4: Back Substitute the Original Variable

Finally, revert back to the original variable \( t \) by substituting \( u = 2t \) back into the expression. This gives the indefinite integral \( \frac{1}{2} e^{2t} + C \).

Key Concepts

Exponential Functions

Exponential functions are a cornerstone in mathematics, characterized by the equation \( y = a^{x} \), where \( a \) is a positive constant. Specifically, the base \( e \) (approximately 2.71828) is frequently used in calculus. The function \( e^x \) demonstrates a unique property: its derivative is the same as the function itself. This self-replicating property under differentiation makes it an excellent candidate for integration problems.
When dealing with exponential functions in integration, as in \( \int e^{2t} \, dt \), one often encounters exponents that are linear in the variable, like \( 2t \).
Recognizing this form is essential as it indicates that substitution might simplify the process. Keep in mind:

• Look for constants multiplied by the variable in exponents (e.g., \( 2t \)).
• Use substitutions to transform the integral into a simpler form when necessary.

This understanding lays the groundwork for effectively applying the substitution method.

Substitution Method

The substitution method is a powerful tool to simplify integrals, especially when dealing with complex functions. The key idea is to replace a part of the integral with a single variable, making it easier to solve. Let's go through this step-by-step:
Consider the integral \( \int e^{2t} \, dt \). Here, the exponent \( 2t \) complicates direct integration. By letting \( u = 2t \), we simplify the expression.
Next, compute the differential: if \( u = 2t \), then \( du = 2 \, dt \). Solving for \( dt \), we have \( dt = \frac{1}{2} \, du \).
Substituting these into the original integral gives \( \int e^u \cdot \frac{1}{2} \, du \). This transformation turns the complex original form into a more manageable problem.

• The substitution should match the form of the integral to a new variable conveniently.
• This method often involves identifying parts of the function derived from a simpler, familiar form like \( e^x \).

After solving the new integral, don't forget the back substitution, returning to the original variable.

Constant of Integration

In indefinite integrals, the solution is usually not a single function but a family of functions. This is indicated by the constant of integration, often denoted as \( C \).
When solving integrals, like \( \int e^{2t} \, dt \), the result is \( \frac{1}{2} e^{2t} + C \), where \( C \) represents an arbitrary constant. This constant arises because antiderivatives are not unique; they differ by a constant when the function is constant over an interval.
Here’s why it's important:

• Integrals contain an unknown constant because differentiation loses this constant information (the derivative of a constant is zero).
• \( C \) ensures all possible antiderivatives of a function are represented.

Never overlook the constant of integration, as it's an integral part of finding the complete set of solutions to an anti-differentiation problem.