Problem 52 - Essential Calculus

Question

Evaluate the integral. \(\int_{0}^{\ln 2} 2 e^{-x} \cosh x d x\)

Step-by-Step Solution

Answer

The value of the integral is \( \ln2 +\frac{3}{8} \).

Step 1: Identify function form

The given integral, \( \int_{0}^{\ln 2} 2e^{-x} \cosh x dx \), involves the product of an exponential function and a hyperbolic cosine function.

Step 2: Recognize the hyperbolic cosine

The hyperbolic cosine function \( \cosh x \) can be expressed in terms of exponential functions as \( \cosh x = \frac{e^{x} + e^{-x}}{2} \)

Step 3: Substitute and simplify

Substitute \(\cosh x\) in this form into the integral. The result is \( \int_{0}^{\ln 2} 2e^{-x} \left(\frac{e^{x} + e^{-x}}{02}\right) dx \therefore \int_{0}^{\ln 2} e^{-x +x} + e^{-x - x} dx \), which simplifies to \( \int_{0}^{\ln 2} e^{0} + e^{-2x} dx = \int_{0}^{\ln 2} 1 + e^{-2x} dx \)

Step 4: Compute the integral

To compute this integral, use the power rule for the first term and the rule for integrating an exponential function for the second term: \( \left[ x -\frac{1}{2}e^{-2x} \right]_{0}^{\ln 2} \)

Step 5: Evaluate the Definite Integral

Substitute the values of the limits \( \ln2 \) and \(0\) to evaluate the integral: \( \left[ (\ln2 -\frac{1}{2}e^{-2 \ln 2} ) - (0 -\frac{1}{2}e^{0}) \right] = \ln2 -\frac{1}{2}e^{-2 \ln 2} +\frac{1}{2} \)

Step 6: Simplify Result

Simplify the result \( \ln2 -\frac{1}{2}e^{-2 \ln 2} +\frac{1}{2}= \ln2 -\frac{1}{2} \cdot \frac{1}{2^2} +\frac{1}{2}= \ln2 -\frac{1}{8} +\frac{1}{2}= \ln2 +\frac{3}{8}\)

Key Concepts

Exponential Functions

Exponential functions are mathematical functions of the form \( f(x) = a^x \) where \( a \) is a constant greater than zero, and \( x \) is a variable. The most common exponential function is the natural exponential function, which uses the constant \( e \) (approximately 2.718) as its base.

Exponential functions are essential in calculus because they model growth and decay processes, such as population growth or radioactive decay. Their unique property is that their rate of change is proportional to their value, making them exceptionally useful for applications across various fields.

• **Base \( e \)**: The base \( e \) leads to the function \( e^x \), known for its natural growth.
• **Properties**: Exponential functions have a constant ratio; for any equal interval on the x-axis, the ratio of the function values remains constant.
• **Integration and Differentiation**: The derivative of \( e^x \) is \( e^x \), and its integral is also \( e^x + C \), where \( C \) is a constant.

In the exercise, the function \( e^{-x} \) is part of the integrand, showcasing the decay nature of the exponential function as \( x \) becomes larger.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are based on hyperbolas instead of circles. Key hyperbolic functions include sinh (hyperbolic sine) and cosh (hyperbolic cosine). They have various applications in engineering, physics, and mathematics due to their natural appearance in hyperboloid structures and relativistic physics applications.

The hyperbolic cosine function, \( \cosh x \), is defined as:\[\cosh x = \frac{e^x + e^{-x}}{2}\]

• **Symmetry**: \( \cosh(x) \) is even, meaning \( \cosh(-x) = \cosh(x) \).
• **Derivative**: The derivative of \( \cosh x \) is \( \sinh x \), providing simple relationships for calculus.

In the given integral, \( \cosh x \) is rewritten using its exponential form. This allows the equation to be simplified easily by directly integrating each exponential term.

Definite Integrals

Definite integrals calculate the area under a curve over a specific interval, represented as \( \int_{a}^{b} f(x) \ dx \). They provide the total accumulation of quantities, such as total distance traveled or net area between curves.

Key aspects of definite integrals are:

• **Limits of Integration**: The values \( a \) and \( b \) specify the interval over which the function is integrated.
• **Evaluation**: After finding the indefinite integral or antiderivative \( F(x) \), substitute the limits of integration to compute \( F(b) - F(a) \).
• **Fundamental Theorem of Calculus**: It connects differentiation and integration, ensuring that differentiation and integration are inverse processes.

In this exercise, the definite integral \( \int_{0}^{\ln 2} \) means calculating the area from \( x = 0 \) to \( x = \ln 2 \). Integration steps show how to simplify components to evaluate the integral efficiently. Eventually, this results in a concrete value representing the accumulated quantity over that interval.