Problem 38 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^1_0 \cosh t \,dt \)

Step-by-Step Solution

Answer

The integral is \( \frac{e - \frac{1}{e}}{2} \).

Step 1: Identify the Function for Integration

We need to evaluate the integral of \( \cosh t \) from 0 to 1. The function \( \cosh t \) is the hyperbolic cosine function, which can be expressed as \( \cosh t = \frac{e^t + e^{-t}}{2} \).

Step 2: Find the Antiderivative

To integrate \( \cosh t \), we find its antiderivative. The antiderivative of \( \cosh t \) is \( \sinh t \) because the derivative of \( \sinh t \) is \( \cosh t \).

Step 3: Evaluate the Indefinite Integral

The indefinite integral of \( \cosh t \) is \( \sinh t + C \), where \( C \) is the constant of integration.

Step 4: Apply the Fundamental Theorem of Calculus

To evaluate the definite integral from 0 to 1, apply the Fundamental Theorem of Calculus: Evaluate \( \sinh t \) at the upper limit 1 and the lower limit 0:\[ \int_0^1 \cosh t \, dt = \left[\sinh t \right]_0^1 = \sinh(1) - \sinh(0) \].

Step 5: Calculate the Values

Now, calculate \( \sinh(1) \) and \( \sinh(0) \). We know \( \sinh t = \frac{e^t - e^{-t}}{2} \).\[\sinh(1) = \frac{e^1 - e^{-1}}{2} = \frac{e - \frac{1}{e}}{2}\]\[\sinh(0) = \frac{e^0 - e^{0}}{2} = 0 \]

Step 6: Subtract the Lower Limit Result

Now, substitute these values back:\[\int_0^1 \cosh t \, dt = \left(\frac{e - \frac{1}{e}}{2} \right) - 0 = \frac{e - \frac{1}{e}}{2}\]

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions that resemble the trigonometric functions but are based on hyperbolas rather than circles. A key hyperbolic function is the hyperbolic cosine, denoted as \( \cosh t \). The hyperbolic cosine function is defined as:

• \( \cosh t = \frac{e^t + e^{-t}}{2} \)

This function takes any real number as input and outputs a corresponding value. Like the cosine function in trigonometry, \( \cosh t \) is an even function, meaning it is symmetrical around the y-axis. This is visible when graphing \( \cosh t \) as it creates a curve that gets closer to the y-axis without ever reaching zero, reflecting its exponential nature.
In calculus, hyperbolic functions often appear in various integration and differentiation problems. Recognizing when a function can be rewritten with hyperbolic identities can simplify the problem-solving process.

Antiderivative

The antiderivative, also known as the indefinite integral, reverses the process of differentiation. It answers the question: "What function has this derivative?"
In this exercise, we seek the antiderivative of the hyperbolic cosine function \( \cosh t \). The antiderivative of \( \cosh t \) is the hyperbolic sine function \( \sinh t \).

• The derivative of \( \sinh t \) is \( \cosh t \).
• Therefore, the indefinite integral of \( \cosh t \) is expressed as \( \sinh t + C \), where C is the constant of integration.

Antiderivatives are essential in calculus as they allow the evaluation of areas under curves and solving of various real-world problems.

Definite Integral

The definite integral computes the accumulation of quantities, often realized as the area under a curve between two bounds. When evaluating a definite integral, such as \( \int_0^1 \cosh t \, dt \), you're finding the total accumulation of the function \( \cosh t \) from \( t = 0 \) to \( t = 1 \).
The process involves computing the antiderivative of the function and applying the following steps:

• Find the antiderivative (here, it's \( \sinh t \)).
• Evaluate it at the upper bound (\( \sinh(1) \)).
• Evaluate it at the lower bound (\( \sinh(0) \)).
• Finally, subtract the lower bound result from the upper bound result: \( \sinh(1) - \sinh(0) \).

This gives the precise value representing the area under the curve between these bounds, highlighting the actual use of integrals in practical scenarios.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a vital concept in integral calculus, creating a powerful link between differentiation and integration. Its essence is captured in two main parts:

• First, it states that if a function is continuous over an interval, then the function has an antiderivative. Thus, you can find the integral of the function using its antiderivative.
• Second, it allows the evaluation of a definite integral by using the antiderivative: \( F(b) - F(a) \), where \( F \) is the antiderivative of \( f \), and \( a \) and \( b \) are the bounds of the interval.

For our example, this means that the definite integral \( \int_0^1 \cosh t \, dt \) can be evaluated by computing \( \left[ \sinh t \right]_0^1 \), resulting in \( \sinh(1) - \sinh(0) \). This theorem beautifully showcases how derivatives and integrals are inversely related, making it a cornerstone of calculus learning.