Problem 25 - Applied Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify Parts for Integration by Parts

For the integration by parts method, we use the formula \( \int u \, dv = uv - \int v \, du \). Start by choosing \( u = x \) and \( dv = \frac{1}{e^{2x}}dx = e^{-2x}dx \). This choice simplifies the problem since the derivative of \( x \) and the integral of \( e^{-2x} \) are straightforward.

Step 2: Differentiate and Integrate the Parts

Differentiate \( u \) and integrate \( dv \): \( du = dx \) and integrate \( dv \) to find \( v \). Since \( dv = e^{-2x} dx \), integrating gives \( v = -\frac{1}{2} e^{-2x} \).

Step 3: Substitute into Integration by Parts Formula

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

Step 4: Simplify the Expression

Simplify the expression: \( -\frac{x}{2} e^{-2x} + \frac{1}{2} \int e^{-2x} dx \). Now, focus on the remaining integral \( \int e^{-2x} dx \).

Step 5: Integrate the Remaining Term

Integrate \( \int e^{-2x} dx \) to find \(-\frac{1}{2} e^{-2x} \) (the antiderivative of \( e^{-2x} \) is \(-\frac{1}{2} e^{-2x} + C \), because of the presence of \(-2\) as a coefficient in the exponent).

Step 6: Combine the Results

Combine the results: \(-\frac{x}{2} e^{-2x} + \frac{1}{2} \left(-\frac{1}{2} e^{-2x}\right) = -\frac{x}{2} e^{-2x} - \frac{1}{4} e^{-2x} + C \). Factor out \( -\frac{1}{4} e^{-2x} \) to consolidate: \(-\frac{1}{4} e^{-2x} (2x + 1) + C\).

Step 7: Final Answer with Constant of Integration

The final expression for the integral is: \(-\frac{1}{4} (2x + 1) e^{-2x} + C\), where \( C \) is the constant of integration.

Key Concepts

Definite Integrals

Definite integrals refer to integrals evaluated over a specific interval. Unlike indefinite integrals, which include a constant of integration, definite integrals have two boundaries, often denoted at the lower and upper limits of integration. Evaluating these integrals requires

• Calculating the antiderivative of the function involved
• Using the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects derivatives with integrals and gives us a way to evaluate definite integrals.

When you compute a definite integral, you're essentially calculating the net area between the function's graph and the x-axis over the specified interval. The limits of integration can either give you a positive area or indicate an overall movement below the x-axis.

In our exercise, we didn't evaluate a definite integral, as it focused on indefinite integrals using integration by parts. However, grasping definite integrals is crucial when applying integration by parts in real-world scenarios involving specific boundaries.

Antiderivatives

Antiderivatives, sometimes known as indefinite integrals, represent a family of functions. They're the reverse process of differentiation. When finding an antiderivative, you're looking for a function whose derivative gives you the original function.

The antiderivative concept is crucial in calculus because it helps solve areas under curves and reversible processes where differentiation is involved, as seen in other functions or physics problems.

• An antiderivative of a simple function like \( f(x) = x \) is \( \frac{x^2}{2} + C \)
• The antiderivative of \( e^x \) is \( e^x + C \).

The constant \( C \) symbolizes an infinite number of potential solutions, which offset varied boundary conditions in application problems.

In our solution, the antiderivative process facilitated the step-by-step integration using parts, particularly the antiderivative \(-\frac{1}{2} e^{-2x} + C\).

This function becomes part of the integration process, pivotal for unraveling complex integration problems.

Integration Techniques

At times, finding integrals directly can be daunting, needing diverse methods or techniques to simplify and solve them effectively. The integration by parts technique comes in handy when dealing with products of functions.

This technique employs the formula:

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

Choosing which parts of the integrand to set as \( u \) and \( dv \) is crucial.

Preferred choices often include:

• Functions that simplify upon differentiation as \( u \), like polynomial functions such as \( x \)
• Functions with straightforward integration as \( dv \), like exponential functions \( e^{ax} \)

Our exercise uses these choices, simplifying the integration process and effectively breaking the integral into more manageable parts. Understanding how to strategically apply various integration techniques is essential, particularly as you encounter more complex integrals in advanced studies.

In the exercise, the integration by parts allowed us to tackle \( \int \frac{x}{e^{2x}} dx \) by reducing it to manageable terms, revealing the result: \(-\frac{1}{4} (2x + 1) e^{-2x} + C\).