Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Evaluate the integrals. $$ \int 3 x e^{-2 x} d x $$

Step-by-Step Solution

Answer

The integral of \(3x e^{-2x}\) dx is \(-\frac{3}{2} x e^{-2x} - \frac{3}{4} e^{-2x} + C\).

Step 1: Identify parts for integration

Firstly, divide the integrand into two parts: \(u\) and \(dv\). Let \(u = x\) and \(dv = 3e^{-2x} dx\). Be aware that the integral now represents the product of these two terms.

Step 2: Utilize formula for integration by parts

Next, the formula for integration by parts \(\int u dv = uv - \int v du\) is applied. To do this, it's needed to find the derivative of \(u\) (denoted as \(du\)) and the integral of \(dv\) (denoted as \(v\)). The derivative \(du\) is simply \(dx\), and \(v\) can be computed by evaluating the integral of \(3e^{-2x}\), which is \(-\frac{3}{2} e^{-2x}\).

Step 3: Implement the formula

Having \(u\), \(v\), \(du\), and \(dv\), substitute them into the formula \(\int u dv = uv - \int v du\). This gives: \(x \cdot -\frac{3}{2} e^{-2x} - \int -\frac{3}{2} e^{-2x} dx\), where the second part, \(\int -\frac{3}{2} e^{-2x} dx\), is now a simpler integral to solve.

Step 4: Solve the remaining integral

Evaluate the integral \(\int -\frac{3}{2} e^{-2x} dx\). This is a basic integral and its result is \(\frac{3}{4} e^{-2x}\).

Step 5: Combine the results

Finally, combine the result of the remaining integral with the other parts of the equation to get the answer. Don't forget to add the constant of integration \(C\), which is needed when computing indefinite integrals. The result is \(-\frac{3}{2} x e^{-2x} - \frac{3}{4} e^{-2x} + C\).

Key Concepts

Indefinite Integrals

Indefinite integrals are a fundamental component of calculus. When we talk about indefinite integrals, we're referring to the process of finding a function, often called an antiderivative, whose derivative is the given function. This process is essentially the reverse of differentiation.

Unlike definite integrals, indefinite integrals do not have upper or lower limits of integration and therefore represent a family of functions. Each function in this family differs by a constant, noted as the constant of integration, represented by the letter "C".

• Symbolically, the indefinite integral is represented as \( \int f(x) \, dx \).
• When integrating, always remember to add \( C \), indicating an arbitrary constant. This is crucial because when derivatives are taken, constants disappear.

The indefinite integral of the function presents a broader spectrum of solutions, each shifting vertically on a graph due to the constant. This property of indefinite integrals is vital when solving problems in physics, engineering, and beyond.

Integration Techniques

Integration techniques provide various methods for tackling integrals that are not straightforward. One common and powerful technique is integration by parts. This method is particularly useful when dealing with products of functions.

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). Here's how it works:

• Identify parts of the integral as \( u \) and \( dv \) such that their combination simplifies the integral.
• Calculate the derivative \( du \) and find \( v \) by integrating \( dv \).
• Substitute these into the formula, which typically transforms the integrand into a more manageable form.

In our example, where we integrated \( \int 3x e^{-2x} \, dx \), integration by parts was helpful because it involved a product of a polynomial \( x \) and an exponential function \( e^{-2x} \). Through this approach, we decomposed a complex integral into a simpler form and ultimately found the integral of the product.

Exponential Functions

Exponential functions are functions in which a constant base is raised to a variable exponent. The exponential function commonly takes the form \( e^x \), where \( e \) is approximately 2.71828. In calculus, exponential functions are unique because their derivative and integral are proportionally related to the function itself.

When dealing with the integration of exponential functions, especially in the form \( e^{ax} \), remember that:

• The integral of \( e^{ax} \) is \( \frac{1}{a} e^{ax} + C \).
• Exponential decay, such as \( e^{-2x} \), implies the function decreases as \( x \) increases.

In the provided exercise, the term \( e^{-2x} \) was integrated using this property. The result is inherently related to how exponential functions behave algebraically and in calculus. Understanding this can simplify the integration of other exponential forms and helps grasp their significance in mathematical modeling.