Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-36, use integration by parts to evaluate each integral. $$ \int x e^{3 x} d x $$

Step-by-Step Solution

Answer

\( \int x e^{3x} \, dx = \frac{1}{3} e^{3x} (x - \frac{1}{3}) + C \).

Step 1: Identify Parts

To use integration by parts, we first need to identify a part to differentiate \( u \) and a part to integrate \( dv \). For this integral \( \int x e^{3x} \, dx \), let \( u = x \) which gives \( du = dx \). Let \( dv = e^{3x} \, dx \), which makes \( v = \frac{1}{3} e^{3x} \) upon integrating \( dv \).

Step 2: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Plugging in the identified parts, we get: \[ \int x e^{3x} \, dx = x \cdot \frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} \cdot dx \].

Step 3: Simplify and Solve Remaining Integral

Simplify the product \( x \cdot \frac{1}{3} e^{3x} \) to \( \frac{x}{3} e^{3x} \). The remaining integral becomes \( -\frac{1}{3} \int e^{3x} \, dx \). Integrating \( e^{3x} \) gives \( \frac{1}{3} e^{3x} \) again, so we have: \[ \int x e^{3x} \, dx = \frac{x}{3} e^{3x} - \frac{1}{3} \cdot \frac{1}{3} e^{3x} + C \].

Step 4: Combine Terms and Simplify

Combine and simplify the terms: \( \frac{x}{3} e^{3x} - \frac{1}{9} e^{3x} \). Factor out \( \frac{1}{3} e^{3x} \) to get: \[ \int x e^{3x} \, dx = \frac{1}{3} e^{3x} \left( x - \frac{1}{3} \right) + C \].

Key Concepts

Definite Integrals

Definite integrals are an essential concept in calculus. They represent the evaluation of the integral of a function over a specific interval, effectively capturing the area under the curve within those bounds. The notation for a definite integral from a to b is written as:

• \( \int_{a}^{b} f(x) \, dx \)

This expression computes the net signed area between the function and the x-axis, from x = a to x = b.To evaluate definite integrals, it's crucial to first find the antiderivative, often called an indefinite integral, and then apply the limits of integration. This process is formalized as the Fundamental Theorem of Calculus, which states that if \( F(x) \) is an antiderivative of \( f(x) \), then:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

Definite integrals are used in various applications such as finding the total distance traveled by an object or the total amount of a substance within a specified interval. Understanding them lays the groundwork for many practical calculus problems.

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, are core to solving various calculus problems. They represent a family of functions and are primarily used to reverse the process of differentiation.The notation for an indefinite integral looks like this:

• \( \int f(x) \, dx \)

The result is a function \( F(x) \) plus a constant of integration \( C \), because differentiation of constants results in zero, leaving multiple potential solutions. Key points to remember about indefinite integrals:

• They lack specific boundaries, unlike definite integrals.
• The process allows you to find a general form of antiderivative for functions like polynomials, exponentials, and trigonometric functions.
• Integration techniques like substitution and integration by parts are often applied to solve them.

By mastering indefinite integrals, you can effectively handle a variety of calculus problems, enabling you to move onto more complex integration techniques.

Integration Techniques

Integration techniques are strategies devised to simplify the process of integrating functions that are not easily integrable in their given form.One such technique is integration by parts, which is particularly useful for integrating products of functions, similar to factoring or distributing in algebra.The integration by parts formula is given by:

• \( \int u \, dv = uv - \int v \, du \)

This technique is crucial when dealing with complex functions. To apply it successfully, correctly identifying parts of the function as \( u \) and \( dv \) is a crucial initial step. Other common techniques include:

• Substitution: Simplifies integrals using a change of variables.
• Partial Fraction Decomposition: Useful for rational functions.

Learning integration techniques allows for solving broader and increasingly complex calculus problems, whether they involve polynomials, exponentials, or trigonometric functions.

Calculus Problems

Calculus problems often involve the application of integrals to model and solve real-world situations. By breaking down problems into understandable parts, calculus further becomes not only manageable but also applicable to practical scenarios. Typical calculus problems include:

• Finding the area under curves to determine distances or consumption levels.
• Calculating rates of change and understanding dynamic systems, such as population growth or chemical reactions.
• Applying calculus techniques to physics problems, like motion under gravity or the flow of fluids.

To effectively tackle calculus problems:

• Clearly understand the problem statement and identify key variables.
• Select the appropriate integration strategy, be it a definite or indefinite integral.
• Break down the solution into smaller steps for better understanding and management.

By mastering the art of solving calculus problems, students can bridge complex mathematical theories with real-life applications, building strong analytical and problem-solving skills.