91Ó°ÊÓ

The substitution method in integral calculus is a powerful tool for simplifying complex integrals. It involves replacing a complicated part of an integral with a new variable, often making the integral easier to solve. In this case, we had the integral \( \int \frac{3 e}{4 + e^{2t}} \, dt \). The exponential expression within the denominator suggests that substitution can reduce complexity. Here's how substitution works:

• We identify a part of the function to substitute with a new variable, \( u \).
• We set \( u = 4 + e^{2t} \), chosen to simplify the denominator's expression.
• Then, calculate the derivative \( du \) in terms of \( dt \). Here, \( du = 2e^{2t} \ dt \).

By substituting, the given function transforms, making the integration process more manageable. The differential \( dt \) can be rewritten using the relation \( dt = \frac{du}{2(u - 4)} \). This conversion simplifies the function into a more straightforward fraction, ready for further methods like partial fractions.

Partial fraction decomposition is a technique used to break complex rational functions into simpler fractions that are easier to integrate. Once we used the substitution method, we transformed our integral into a new form: \[ \frac{3e}{2} \int \frac{1}{(u-4)u} \, du \].This expression is a rational function, where partial fraction decomposition becomes quite useful. Here's how we apply it:

• Express \( \frac{1}{(u-4)u} \) as a sum of two separate fractions: \( \frac{A}{u} + \frac{B}{u-4} \).
• The goal is to find constants \( A \) and \( B \).
• By equating and simplifying the terms, we derive \( A = \frac{1}{4} \) and \( B = -\frac{1}{4} \).

Now, the integral can easily be addressed as two separate, simpler integrals:\[ \frac{3e}{2} \left( \int \frac{1}{4u} \, du - \int \frac{1}{4(u-4)} \, du \right). \]Each portion can be integrated on its own, leading to a solution that can be reassembled into a single expression.

Exponential functions, such as \( e^{2t} \) in this problem, often appear in calculus due to their common occurrence in growth and decay models. They are defined as functions of the form \( e^{x} \), where \( e \) is Euler's number (approximately 2.71828), a fundamental mathematical constant. In the context of solving integrals, working with exponential functions involves recognizing their properties. For instance:

• Derivative: \( \frac{d}{dt}[e^{kt}] = ke^{kt} \).
• Integral: \( \int e^{kt} \, dt = \frac{1}{k} e^{kt} + C \).

In our exercise, \( e^{2t} \) appeared in the exponent, suggesting an exponential growth context. Utilizing substitution helped simplify the compounding nature of exponential terms. Once substituted into a simpler form, this enabled us to leverage easier algebraic manipulation techniques. Integrating exponential expressions requires recognizing these key characteristics, allowing us to handle even complex integrals effortlessly by transforming them into manageable components.