Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-36, use integration by parts to evaluate each integral. $$ \int x 2^{x} d x $$

Step-by-Step Solution

Answer

\( \int x \, 2^x \, dx = \frac{x \, 2^x}{\ln(2)} - \frac{2^x}{(\ln(2))^2} + C \).

Step 1: Identify the Functions for Integration by Parts

For integration by parts, we need to choose which parts of the integrand will be differentiated and integrated. Integration by parts is based on the formula \[ \int u \, dv = uv - \int v \, du \].We choose \( u = x \) (which upon differentiation becomes simpler) and \( dv = 2^x \, dx \) (which is easily integrable).

Step 2: Differentiate and Integrate Selected Functions

Differentiate \( u = x \) to find \( du \). Thus, \( du = dx \).Integrate \( dv = 2^x \, dx \) to get \( v = \frac{2^x}{\ln(2)} \), because \( \int 2^x \, dx = \frac{2^x}{\ln(2)} \).

Step 3: Apply the Integration by Parts Formula

Now apply the formula \( \int u \, dv = uv - \int v \, du \).Substituting the values, we get:\[ \int x \, 2^x \, dx = x \cdot \frac{2^x}{\ln(2)} - \int \frac{2^x}{\ln(2)} \, dx \].

Step 4: Simplify and Evaluate the Remaining Integral

The integral \( \int \frac{2^x}{\ln(2)} \, dx \) is similar to the integration in Step 2:\[ \int \frac{2^x}{\ln(2)} \, dx = \frac{1}{\ln(2)} \cdot \frac{2^x}{\ln(2)} = \frac{2^x}{(\ln(2))^2} \].Thus, our expression becomes:\[ x \cdot \frac{2^x}{\ln(2)} - \frac{2^x}{(\ln(2))^2} \].

Step 5: Write the Final Integrated Solution

Including the constant of integration \( C \), the evaluated integral is:\[ \int x \, 2^x \, dx = \frac{x \, 2^x}{\ln(2)} - \frac{2^x}{(\ln(2))^2} + C \].

Key Concepts

Definite Integrals

Definite integrals are a concept that comes into play when we need to calculate the exact area under a curve between two specified points, known as the bounds. Unlike indefinite integrals which include a constant of integration, definite integrals calculate a specific value.
This value represents the total accumulated amount between the lower and the upper limits on the graph of the function.
Understanding the limits is crucial:

• The lower limit is the starting point on the x-axis.
• The upper limit marks the ending point.

When calculating definite integrals, use the Fundamental Theorem of Calculus. This theorem connects the concept of differentiation and integration by stating that if a function is continuous over an interval, then the definite integral over that interval can be found using its antiderivative.
For our exercise, although we solved an indefinite integral, this foundational concept also applies if evaluating between certain bounds to find a specific area.

Differentiation

Differentiation is a key concept that serves as the foundation for many calculus operations, such as integration by parts. It involves finding the derivative of a function, which represents the rate of change of the function with respect to a variable.
In this exercise, we begin by differentiating the function we choose for "u". Here, with the choice of \( u = x \), differentiating gives us \( du = dx \). This is straightforward because the derivative of \( x \) with respect to \( x \) is 1.
In terms of its importance:

• Differentiation helps simplify equations, making integral evaluation more feasible.
• It is used to find slopes of tangents to curves.
• An understanding of derivatives supports problem-solving in physics, engineering, and other fields.

Without differentiation, it would be challenging to progress in many calculus-based solutions, especially those that involve multiple components like integration by parts.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent, commonly denoted as \( b^x \). These functions grow rapidly and are fundamental in various fields like science and finance.
The base \( b \) is often \( e \) (the natural base of logarithms) or another constant like 2, as seen in our integration exercise with \( 2^x \).
Key characteristics of exponential functions include:

• Their increasing or decreasing nature depending on the base.
• Applications in modeling population growth, radioactive decay, or compound interest.

When integrating exponential functions, notice how they often retain their form. For instance, integrating \( 2^x \) results in \( \frac{2^x}{\ln(2)} \). The inclusion of the natural logarithm \( \ln(2) \) is necessary because it adjusts the rate of growth embodied by the expression. Understanding these nuances aids in both differentiation and integration processes of exponential functions.

Mathematical Solutions

Mathematical solutions refer to the comprehensive methods used to solve equations systematically, ensuring accuracy and consistency. The goal is to arrive at a correct solution that addresses the problem's requirements.
In calculus, especially within problems involving integration by parts, mathematical solutions employ a blend of techniques to simplify and solve complex integrals.
For example, in our problem, we:

• Selected appropriate functions for differentiation and integration (choosing \( u = x \) and \( dv = 2^x \, dx \)).
• Applied the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).
• Simplified and combined terms to find the integral in its simplest form.

The Role of Constants

The inclusion of constants, like \( C \) in indefinite integrals, acknowledges the variety of potential antiderivatives. Ensuring you follow all these steps effectively is key to achieving precise mathematical solutions.