Problem 9 - Applied Calculus

Question

Use integration by parts to find each integral. \(x e^{2 x} d x\)

Step-by-Step Solution

Answer

\( \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C \).

Step 1: Identify Parts for Integration by Parts

Integration by parts is based on the formula: \[ \int u \, dv = uv - \int v \, du \]For the integral \( \int x e^{2x} \, dx \), we can choose: - \( u = x \) - \( dv = e^{2x} \, dx \)

Step 2: Differentiate and Integrate the Parts

First, we differentiate \( u \): \[ du = dx \]Next, we integrate \( dv \): \[ v = \int e^{2x} \, dx = \frac{1}{2}e^{2x} \]This integration comes from the rule for integrating \( e^{ax} \).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int x e^{2x} \, dx = uv - \int v \, du \]This becomes:\[ x \left(\frac{1}{2}e^{2x}\right) - \int \frac{1}{2}e^{2x} \, dx \]

Step 4: Simplify and Integrate Remaining Integral

Simplify the expression: \[ \frac{1}{2}x e^{2x} - \frac{1}{2} \int e^{2x} \, dx \]The remaining integral is: \[ \frac{1}{2} \int e^{2x} \, dx = \frac{1}{2} \times \frac{1}{2}e^{2x} = \frac{1}{4}e^{2x} \]

Step 5: Combine Terms and Include Constant of Integration

Putting it all together:\[ \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C \]

Key Concepts

Definite Integrals

When discussing integrals, it's important to understand the concept of definite integrals. Unlike indefinite integrals, which provide a general form of antiderivatives, definite integrals compute the area under a curve over a specified interval. This interval is defined by its upper and lower limits, often denoted as \( a \) and \( b \).
The definite integral of a function \( f(x) \) from \( a \) to \( b \) is written as:

• \[ \int_{a}^{b} f(x) \, dx \]

This represents the net area between the function \( f(x) \) and the x-axis, from \( x = a \) to \( x = b \). If the function is above the x-axis, the area is positive, and if below, it's negative.
Definite integrals are essential in various fields such as physics and engineering, as they help in calculating quantities like distance, area, and volume. While the exercise given pertains to an indefinite integral (since no specific limits were provided), understanding definite integrals is valuable for broader applications.

Exponential Functions

Exponential functions are crucial to understanding calculus and integration problems like the one given in the exercise. An exponential function has the form \( f(x) = a^{x} \), or more commonly in calculus, \( e^{x} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718.

• Exponential functions grow rapidly and are characterized by their constant rate of growth proportional to their size.
• They are defined for all real numbers and are always positive.

The derivative and integral of exponential functions are often simple. For instance, the derivative of \( e^{2x} \) is \( 2e^{2x} \), whereas its integral is \( \frac{1}{2}e^{2x} \), as highlighted in the solution.
An important thing to note is the substitution method, which helps simplify the integration of exponential functions. The problem uses this method by setting \( dv = e^{2x} \, dx \) and then integrating \( dv \) to find \( v \). This technique leverages the inverse nature of differentiation and integration of exponential functions.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative represents the rate of change or the slope of the function at any given point. For the function \( u = x \) in our exercise, differentiating gives \( du = dx \).

• The power rule, a basic differentiation rule, states that the derivative of \( x^n \) is \( nx^{n-1} \).
• Differentiation helps in finding functions' behaviors, like increasing or decreasing trends.

In the integration by parts method, which the exercise explores, differentiation is used to find \( du \) after selecting \( u \). This process is vital as it simplifies integrating the product of two functions.
Differentiation not only aids in computing integrals through methods like integration by parts, but it's also essential in solving real-world problems involving motion, growth, and decay, where rates of change are critical.