Problem 29 - Calculus: Early Transcendental Functions

Question

Evaluate the integral. $$\int_{0}^{1} \frac{e^{x}-1}{e^{2 x}} d x$$

Step-by-Step Solution

Answer

The value of the integral is \(-\frac{1}{e} - \frac{1}{2e^2} + 1.5\)

Step 1: Simplify the Integral

Notice that the integral can be split into two separate integrals: \( \int_{0}^{1} (e^{-x} - e^{-2x}) dx \)

Step 2: Evaluate the Integral of Each part

The integral of \(e^{-x}\) from 0 to 1 is -1 (since the integral of \(e^{-x}\) is \(-e^{-x}\)), and the integral of \(e^{-2x}\) from 0 to 1 is 1/2 (since the integral of \(e^{-2x}\) is \(-e^{-2x}/2\)). Therefore, the original integral equals \([-e^{-x} - \frac{e^{-2x}}{2} ]_0^1\).

Step 3: Evaluate at the bounds

Substituting the bounds into the expression gives \(-e^{-1} - \frac{e^{-2}}{2} + 1 + 0.5 = -\frac{1}{e} - \frac{1}{2e^2} + 1.5\)

Key Concepts

Definite Integral

In Integral Calculus, a definite integral refers to the evaluation of a function's integral over a given interval. It allows us to find the total accumulation of quantities, like area under a curve, across specified bounds.
The definite integral \( \int_{a}^{b} f(x) \, dx \) calculates the net area bounded by the graph of \( f(x) \) between \( x = a \) and \( x = b \).

• \( a \) and \( b \) are the limits of integration, representing the interval.
• The result combines positive and negative areas where the function is above or below the x-axis.

For example, in the problem, the integral \( \int_{0}^{1} \frac{e^{x}-1}{e^{2 x}} \, dx \) is evaluated between \( x = 0 \) and \( x = 1 \), enabling us to find the net area under the curve within these limits. This provides a specific numerical result representing how the function behaves across this range.
Typically, the Fundamental Theorem of Calculus is used, which connects integrals and derivatives, facilitating the computation of these integrals.

Integration Techniques

Integration techniques are methods used to solve integrals, especially when dealing with complex functions. Many types of functions require different approaches to find their integral. Here are some essential techniques:

• Simplification: The first step often involves simplifying the integrand. In the given problem, the integrand \( \frac{e^{x}-1}{e^{2 x}} \) can be separated into \( e^{-x} - e^{-2x} \), making it easier to handle.
• Substitution: Useful when the integrand involves a composite function that isn’t directly integrable.
• Integration by Parts: Applies to products of functions, following a rule similar to the product rule in differentiation.
• Partial Fractions: Utilized when the integrand is a rational function, breaking it into simpler fractions.

For our example, noticing that the function can be rewritten into a simpler form was crucial. Breaking it into two simpler functions, \( e^{-x} \) and \( e^{-2x} \), allowed us to evaluate each part separately using the basic integration rules for exponential functions.

Exponential Functions

Exponential functions have the form \( f(x) = a^{x} \), where \( a \) is a positive constant. They exhibit rapid growth or decay and are common in scientific and financial computations. Integration of exponential functions typically follows straightforward rules:

• \( \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C \)
• Negative exponents like \( e^{-x} \) represent decay.

For instance, when solving \( \int_{0}^{1} (e^{-x} - e^{-2x}) \, dx \), the integrals of \( e^{-x} \) and \( e^{-2x} \) are calculated as \( -e^{-x} \) and \( -\frac{1}{2}e^{-2x} \) respectively over the given interval [0, 1].
By understanding these functions' properties and applying simple integration techniques, students can confidently solve integrals involving exponential expressions. The exponential function’s distinctive characteristic – its derivative and integral being proportional to the function itself – significantly simplifies many calculus problems.