Problem 32 - Applied Calculus

Question

Find each integral by using the integral table on the inside back cover. $$ \int \frac{e^{2 t}}{1-e^{t}} d t $$

Step-by-Step Solution

Answer

The integral is \(- \ln |1 - e^t| + 2(1 - e^t) - \frac{(1 - e^t)^2}{2} + C.\)

Step 1: Identifying the Integral Type

The given integral is \( \int \frac{e^{2t}}{1-e^t} \, dt \). By examining the form, we notice that it resembles the form \( \int \frac{e^{at}}{b - e^{ct}} \, dt \), which often requires substitution to simplify.

Step 2: Substitution

To simplify, use the substitution \( u = 1 - e^t \), which implies that \( du = -e^t \, dt \). Therefore, \( e^t \, dt = -du \) and \( e^{2t} = (1-u)^2 \). Substituting these into the integral gives us: \[ \int \frac{(1-u)^2}{u} (-du) = -\int \frac{(1-u)^2}{u} \, du. \]

Step 3: Simplifying the Integral

Distribute \((1-u)^2\) over \(u\) to get \[ - \int \left( \frac{1}{u} - 2 + u \right) \, du. \] This breaks into three separate integrals: \[ - \int \frac{1}{u} \, du + 2 \int du - \int u \, du. \]

Step 4: Evaluating the Integrals

Evaluate each of the integrals separately: - \( - \int \frac{1}{u} \, du = - \ln |u| \),- \( 2 \int du = 2u \), and- \( - \int u \, du = - \frac{u^2}{2}. \)

Step 5: Combining Results

Combine the results from the integrals to obtain:\[ - \ln |u| + 2u - \frac{u^2}{2} + C. \]

Step 6: Back Substitution

Replace \(u\) with the original variable: \( u = 1 - e^t \). Thus, the integral becomes: \[ - \ln |1 - e^t| + 2(1 - e^t) - \frac{(1 - e^t)^2}{2} + C. \] Simplify if possible.

Key Concepts

Integral Table

Integral tables are incredibly handy when evaluating or solving integrals. They provide a collection of standard integral forms and their corresponding results. This is particularly useful for integrals that can have complicated or non-standard forms.

The integral table acts similar to a dictionary for integrals. Instead of solving complex integrals from scratch, you can look up certain standard forms and use the solutions provided.

• Identify the form of your integral by examining its structure.
• Locate the corresponding form in the integral table.
• Apply any listed solutions as a guide to find your integral's solution.

By employing the integral table, you can often save time and reduce the complexity of manually solving an integral. This method comes in handy for both definite and indefinite integrals. For instance, in our exercise, recognizing the integral's type helped us choose the correct substitution, leading to a simpler form of integration.

Substitution Method

The substitution method, often referred to as "u-substitution," is a technique used to simplify integrals into a more solvable form. It is akin to changing variables in order to make the integral more manageable.

• First, identify which part of the integral can be replaced with a new variable \( u \).
• Differentiate \( u \) to find \( du \) and replace the corresponding part of the integral.
• Carry out the integral with respect to the new variable \( u \).

This method is particularly useful for integrals with composite functions or nested operations. In the given solution, we used \( u = 1 - e^t \) to transform the integral. This converted the complex fraction into a simpler polynomial form, which was easier to integrate. After integration, remember to substitute back the original variable to return to the original context of the problem.

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is crucial in calculus. Each serves a different purpose and result in distinct solutions.

• An indefinite integral represents a family of functions and includes a constant \( C \). Writing it as \( \int f(x) \, dx = F(x) + C \) signifies the general antiderivative of the function.
• A definite integral calculates the net area under a curve between two specific points. It is written as \( \int_{a}^{b} f(x) \, dx \) and results in a numerical value, representing the accumulation of the function's values over the interval \( [a, b] \).

In our exercise, we solved an indefinite integral by finding the antiderivative of the function. The result was expressed as a general solution, including the constant \( C \). Knowing when to apply each type of integral is essential for solving calculus problems effectively.