Problem 41 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^4_0 2^s \,ds \)

Step-by-Step Solution

Answer

The evaluated integral is \( \frac{15}{\ln(2)} \).

Step 1: Identify the Form of the Integral

The integral we need to evaluate is \( \int_0^4 2^s \, ds \). This is an exponential function integral of the form \(\int a^x \, dx\).

Step 2: Recall the Formula for Exponential Integrals

The integral of an exponential function \( a^x \) with respect to \( x \) is given by \( \int a^x \, dx = \frac{a^x}{\ln(a)} + C \), where \( C \) is the constant of integration. Note that this formula applies for \( a > 0 \) and \( a eq 1 \).

Step 3: Apply the Formula to the Given Integral

We apply the formula to \( \int 2^s \, ds \). Thus, \( \int 2^s \, ds = \frac{2^s}{\ln(2)} + C \).

Step 4: Evaluate the Definite Integral

Since we have a definite integral \( \int_0^4 2^s \, ds \), evaluate it as \[ \left[ \frac{2^s}{\ln(2)} \right]_0^4 = \frac{2^4}{\ln(2)} - \frac{2^0}{\ln(2)} \].

Step 5: Calculate the Values

Calculate each part: \( 2^4 = 16 \) and \( 2^0 = 1 \). Therefore, the integral becomes \[ \frac{16}{\ln(2)} - \frac{1}{\ln(2)} = \frac{16 - 1}{\ln(2)} \].

Step 6: Simplify the Result

Simplify the arithmetic: \( 16 - 1 = 15 \). So the evaluated integral is \[ \frac{15}{\ln(2)} \].

Key Concepts

Definite Integral

A definite integral, denoted as \( \int_a^b f(x) \, dx \), calculates the area under the curve of a function \( f(x) \) between two limits, \( a \) and \( b \). It gives a numerical value that represents the accumulated quantity between these bounds. Definite integrals are essential in various fields such as physics, engineering, and economics to compute quantities like area, displacement, and total cost.
In our example, the integral \( \int_0^4 2^s \, ds \) is a definite integral calculating the accumulated area under the exponential curve \( 2^s \) from \( s = 0 \) to \( s = 4 \). With definite integrals, after the function is integrated, the values at the bounds are plugged into the integrated function to find the total accumulated change.

• Lower Limit (\(a\)): 0
• Upper Limit (\(b\)): 4
• Area under \(2^s\) from \(s = 0\) to \(s = 4\)

Calculating this integral provides a numerical value that represents this area, which can be crucial in understanding growth patterns, energy states, or other cumulative measures in real-world scenarios.

Exponential Growth

Exponential growth refers to the process where a quantity increases at a rate proportional to its current value. This type of growth is often described by functions of the form \( a^x \), where \( a \) is a positive constant different from 1. Exponential functions rapidly increase, which reflects phenomena such as population growth, radioactive decay, and compound interest.
In our integral \( \int_0^4 2^s \, ds \), the function \( 2^s \) depicts exponential growth, where 2 is the base of the exponential. As \( s \) increases, \( 2^s \) grows exponentially, doubling with every unit increase in \( s \).
Some key aspects of exponential functions include:

• They have a constant rate of growth.
• The base, if greater than 1, results in rapid increases.
• Used to model realistic scenarios where immediate and sustained growth is observed, like viral spread or technological development.
  
• Graphically, exponential functions have a characteristic steep curve.

In integration scenarios, understanding exponential growth is pivotal because it affects both the setup and solution of integrals involving exponential terms.

Integration Techniques

Integration techniques are methods used to find the integral of functions, which may not be straightforward. By understanding specific rules, like power rule or techniques dealing with exponential functions, throwing light on solving complex real-world problems becomes easier.
For exponential functions that lack easy antiderivatives, the integration by formula helps. One important rule for integrals involving exponentials like \( a^x \) is:

• The integral \( \int a^x \, dx = \frac{a^x}{\ln(a)} + C \). This applies when \( a > 0 \) and \( a eq 1 \).

In our exercise, integrating \( 2^s \) utilizes this formula, emphasizing its crucial role in dealing with exponential functions.
When working through problems:

• Identify the form of the function; recognize if it matches known integral forms.
• Apply applicable formulae wisely for efficient solutions.
• Where possible, simplify calculations with basic arithmetic operations after integration.

Mastering these techniques allows tackling more complex integrations efficiently and accurately across various disciplines.