Problem 25 - Calculus and its applications

Question

Evaluate using integration by parts or substitution. (Assume \(u>0\) in \(\ln\) u. Check by differentiating. $$ \int x^{2} e^{2 x} d x $$

Step-by-Step Solution

Answer

The integral of \( \int x^2 e^{2x} \, dx \) is \( \frac{x^2}{2} e^{2x} - \frac{x}{2} e^{2x} + \frac{1}{4} e^{2x} + C \).

Step 1: Choose the Method

The integral \( \int x^2 e^{2x} \, dx \) can be solved using integration by parts. In this case, we will identify parts of the integrand for integration by parts which follows the formula \( \int u \, dv = uv - \int v \, du \).

Step 2: Identify u and dv

Let \( u = x^2 \) and thus \( du = 2x \, dx \). Let \( dv = e^{2x} \, dx \), so we find \( v \) by integration: \( v = \frac{1}{2} e^{2x} \).

Step 3: Apply Integration by Parts Formula

Substituting into the formula \( \int u \, dv = uv - \int v \, du \), we get: \[ \int x^2 e^{2x} \, dx = \left( x^2 \cdot \frac{1}{2} e^{2x} \right) - \int \left( \frac{1}{2} e^{2x} \cdot 2x \right) \, dx \] which simplifies to \[ \frac{x^2}{2} e^{2x} - \int x e^{2x} \, dx \].

Step 4: Solve the Remaining Integral

Now solve \( \int x e^{2x} \). Using integration by parts again, let \( u = x \), \( du = dx \), \( dv = e^{2x} \, dx \), then \( v = \frac{1}{2} e^{2x} \). Substituting, \( \int x e^{2x} \, dx = \left( x \cdot \frac{1}{2} e^{2x} \right) - \int \left( \frac{1}{2} e^{2x} \cdot 1 \right) \, dx \).

Step 5: Simplify the Results

Simplify the expression: \[ \frac{x}{2} e^{2x} - \frac{1}{2} \int e^{2x} \, dx = \frac{x}{2} e^{2x} - \frac{1}{2} \left( \frac{1}{2} e^{2x} \right) \] which results in \( \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} \).

Step 6: Combine Results

Substitute back into the expression from Step 3: \[ \frac{x^2}{2} e^{2x} - \left( \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} \right) \] which simplifies to \( \frac{x^2}{2} e^{2x} - \frac{x}{2} e^{2x} + \frac{1}{4} e^{2x} \).

Step 7: Check by Differentiating

Differentiate \( \frac{x^2}{2} e^{2x} - \frac{x}{2} e^{2x} + \frac{1}{4} e^{2x} \) to verify the solution. Differentiating gives \[ (x^2 e^{2x} - x e^{2x} + \frac{1}{2} e^{2x}) \] which matches the original integrand.

Key Concepts

Indefinite Integrals

Indefinite integrals are a fundamental concept in calculus that represent the antiderivative of a function. They show the most general form of a function whose derivative is given. When you integrate a function without specific limits of integration, the result is called an indefinite integral. This is often expressed as \[\int f(x) \, dx = F(x) + C\]where \(F(x)\) is the antiderivative of \(f(x)\) and \(C\) is the constant of integration.
This constant is important because when you differentiate an antiderivative, the differentiation process removes constants.
In practical terms, indefinite integrals help us find the original function before it was differentiated. This is useful in various applications, such as calculating the area under a curve, solving differential equations, and more.

• The indefinite integral of a function is a family of functions.
• It includes a constant of integration \(C\), because the antiderivative is not unique.
• Can be thought of as "undoing" differentiation.

Understanding indefinite integrals is crucial when studying calculus, as they form the basis for solving many kinds of problems, including those involving differential equations and area calculations.

Differentiation

Differentiation is the process of finding the derivative of a function. This derivative represents the rate at which something is changing. It essentially tells you the slope of a function at any given point. When you differentiate an integral, you are checking that the antiderivative is correct by essentially reversing the integration process.
In the context of this exercise, after solving the integral, differentiation allows us to verify the solution. If you differentiate the result of an integral and arrive back at your original function, your solution is confirmed to be correct.

• Derivatives are used to find the slope of a curve or a rate of change.
• Essential for understanding how quantities change with respect to one another.
• Checking integrals via differentiation ensures the accuracy of calculus operations.

When learning differentiation, one starts with basic rules, such as power, product, quotient, and chain rules. These rules simplify the process of finding derivatives for different types of functions.
This concept is not only crucial in calculus but also widely applicable in physics, engineering, economics, and other fields where change is analyzed.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. They are written in the form \(f(x) = a \, e^{bx}\), where \(e\) is the base of the natural logarithm, approximately equal to 2.718. These functions have unique properties that make them particularly interesting in calculus.
Exponential functions arise often in situations involving growth and decay, such as population growth, radioactive decay, and compound interest.
In the context of integration and differentiation, the exponential function \(e^{x}\) is unique because it is its own derivative. This property simplifies many calculus operations involving exponential functions.

• They model a wide range of real-world phenomena involving exponential growth or decay.
• In calculus, integration and differentiation of exponentials are straightforward due to their unique properties.
• Understanding these functions is essential for tackling many problems in physics and mathematics.

When working with exponential functions in integration by parts, as in this exercise, recognizing their properties helps simplify calculations and results in more straightforward problem-solving.