Problem 42 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the definite integral. $$ \int_{-1}^{1}\left(e^{x}-e^{-x}\right) d x $$

Step-by-Step Solution

Answer

The evaluated definite integral equals \(e-1/e\).

Step 1: Identify the Function Type

In this step, we recognize that the function under the integral sign \(e^{x} - e^{-x}\) corresponds to the hyperbolic function definition. However, for the calculus, it's easier to keep it as the sum of exponential functions.

Step 2: Break Down the Integral

Now, this step involves breaking down the problem into two separate integrals. Instead of calculating \(\int_{-1}^{1} (e^{x}-e^{-x}) dx\), we can actually compute it as \(\int_{-1}^{1} e^{x} dx - \int_{-1}^{1} e^{-x} dx\). This will facilitate our calculation.

Step 3: Evaluate the Definite Integrals

Calculate both integrals separately. The integral of \(e^{x}\) is \(e^{x}\) and the integral of \(e^{-x}\) is \(-e^{-x}\). Substitute the limits in. For \(\int_{-1}^{1} e^{x} dx\), the result is \(e^{1}-e^{-1}\). For \(\int_{-1}^{1} e^{-x} dx\), the result is \(-e^{-1}+e\).

Step 4: Final Calculation

Now sum the results from Step 3. \(e^{1}-e^{-1}-e^{-1}+e=e+1/e-2/e=e-1/e\).

Key Concepts

Exponential Functions

Exponential functions are a fundamental concept in mathematics. They are essential in modeling situations where quantities grow or decay at a constant rate. An exponential function is typically expressed in the form of \( e^{x} \), where \( e \) is a mathematical constant approximately equal to 2.71828. This constant is known as Euler's number.

• The function \( e^{x} \) represents exponential growth when the exponent \( x \) is positive.
• Conversely, \( e^{-x} \) indicates exponential decay, as higher values of \( x \) make the function decrease towards zero.

Understanding exponential functions is crucial in various fields, including biology, physics, and finance. They can describe everything from population growth to radioactive decay. In the context of the given integral, the function \( e^{x} - e^{-x} \) involves exponential functions with opposite signs, which helps represent specific transformations of curve behavior on a graph.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for hyperbolas instead of circles. They arise from exponential functions and are defined using \( e^{x} \) and \( e^{-x} \). Hyperbolic functions include hyperbolic sine (\( \sinh(x) \)), hyperbolic cosine (\( \cosh(x) \)), and others.

• The hyperbolic sine function is \( \sinh(x) = \frac{e^{x} - e^{-x}}{2} \).
• This function is similar in shape to the normal sine function but stretches along the hyperbola.
• The hyperbolic cosine function is \( \cosh(x) = \frac{e^{x} + e^{-x}}{2} \) and resembles the regular cosine function in shape.

When tackling integrals involving terms like \( e^{x} - e^{-x} \), it helps to recognize their connection to hyperbolic functions. This can simplify solving the integral by identifying familiar patterns from hyperbolic identities.

Integral Calculus

Integral calculus is a branch of mathematics that focuses on finding the total accumulation of quantities. It contrasts with differential calculus, which is about rates of change. An integral can be thought of as the area under a curve when graphed. There are two primary types of integrals: indefinite and definite.

• An indefinite integral represents a family of functions and is presented with an integration constant \( +C \).
• A definite integral computes the exact area between a function and the x-axis over a specified interval, providing a numerical value.

For the given problem, a definite integral is used to evaluate \( \int_{-1}^{1}(e^{x} - e^{-x}) \, dx \). This involves calculating separate areas for \( e^{x} \) and \( e^{-x} \) over the interval from -1 to 1:

• Integral of \( e^{x} \) over \([-1, 1]\) gives \( e - \frac{1}{e} \).
• Integral of \( e^{-x} \) results in \( \frac{1}{e} - e \).

Finally, the results are combined to find the total area or value of the definite integral.