Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int x 2^{x} d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify parts for integration by parts

Integration by parts is based on the formula \( \int u \, dv = uv - \int v \, du \). We need to identify \( u \) and \( dv \) for our integral \( \int x \, 2^x \, dx \).Let's choose \( u = x \) and \( dv = 2^x \, dx \).

Step 2: Differentiate and Integrate

Now differentiate \( u \) and integrate \( dv \):- Differentiate \( u = x \) to get \( du = dx \).- Integrate \( dv = 2^x \, dx \). The integral of \( 2^x \) is \( \frac{2^x}{\ln(2)} \), so \( v = \frac{2^x}{\ln(2)} \).

Step 3: Apply the integration by parts formula

Substitute \( u \), \( du \), \( v \) into the integration by parts formula:\[ \int x \, 2^x \, dx = x \cdot \frac{2^x}{\ln(2)} - \int \frac{2^x}{\ln(2)} \, dx \]

Step 4: Evaluate the remaining integral

Now evaluate the remaining integral \( \int \frac{2^x}{\ln(2)} \, dx \):\( \int \frac{2^x}{\ln(2)} \, dx = \frac{1}{\ln(2)} \int 2^x \, dx = \frac{1}{\ln(2)} \cdot \frac{2^x}{\ln(2)} = \frac{2^x}{[\ln(2)]^2} \).

Step 5: Combine results

Combine the results from previous steps:\[ \int x \, 2^x \, dx = x \cdot \frac{2^x}{\ln(2)} - \frac{2^x}{[\ln(2)]^2} + C \]Where \( C \) is the constant of integration.

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change. To understand calculus, imagine you're examining how a car speeds up or slows down. This subject is all about studying such dynamic processes. Calculus can be broken down into two main parts: differentiation and integration.

- **Differentiation** is concerned with rates of change. Think of it as pinpointing how fast something is moving at a specific moment, like reading a speedometer. - **Integration** revolves around accumulating quantities. It can be likened to figuring out the total distance traveled by adding up small segments over time.

By using these concepts, calculus allows us to solve complex problems in physics, engineering, and economics. It's a tool for seeing the "big picture" by understanding the small details.

Integrals

Integrals are central to calculus, representing accumulation of quantities. When you see an integral sign, \(\int\), picture it as a sum of an infinite number of small parts. This idea is crucial for finding areas under curves or solving problems involving total change.

Let's consider the basic types of integrals:

- **Definite Integrals** compute a specific value, such as an area.- **Indefinite Integrals** provide a generalized function plus a constant.

Using integrals enables a deeper look into continuous quantities, like calculating total earnings based on an income rate, or determining the area under varying curves. The beauty of integrals is they let us move from the specific to the broad, giving a comprehensive overview of the matter at hand.

Integration Techniques

Integration techniques are methods used to find the integral of various functions. Each technique is like a tool, tailored for particular types of integrals and situations.

Some common integration techniques include:

- **Substitution Method**, often referred to as "u-substitution", simplifies complex functions by substituting them with a simpler variable. - **Integration by Parts**, which is especially useful when the integrand is a product of two functions, as showcased in the original exercise.

Choosing the right technique depends on the function's form. For example, when dealing with products of polynomial and exponential functions, integration by parts is often the go-to solution. This technique breaks down the integral into simpler parts, making it manageable and straightforward to evaluate.

U-substitution

U-substitution is a fundamental technique in integration, designed to render challenging integrals into simpler ones. Often compared to a change of perspective, it allows us to reframe a problem for easier handling.

Here's how it works: Identify a part of the integrand, which can be substituted with a variable, typically denoted as "u". This substitution simplifies the integral, transforming it into a more straightforward form to tackle.

- **Choose u**: Determine which part of the integrand to substitute.- **Differentiate**: Find \( du \) by differentiating \( u \).- **Replace**: Substitute \( u \) and \( du \) into the integral.

This method is intuitive, allowing us to visually "see" the integral's complexity and isolate it into manageable pieces. While the original exercise didn't call for u-substitution, understanding it deepens one's proficiency in handling diverse integral problems, enabling fluid transitions across techniques.