Problem 6 - Calculus and its applications

Question

Evaluate using integration by parts or substitution. (Assume \(u>0\) in \(\ln\) u. Check by differentiating. $$ \int 2 x e^{4 x} d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{2}xe^{4x} - \frac{1}{8}e^{4x} + C \).

Step 1: Identify the parts for integration by parts

To use integration by parts, we need to identify the functions for which we will apply the formula \( \int u \, dv = uv - \int v \, du \). Let us choose \( u = 2x \) and \( dv = e^{4x} \; dx \). Thus, we will differentiate \( u \) and integrate \( dv \).

Step 2: Differentiate and integrate

Differentiate \( u = 2x \) to get \( du = 2 \, dx \). Integrate \( dv = e^{4x} \, dx \) to get \( v = \frac{1}{4}e^{4x} \). This gives us all the parts needed for integration by parts.

Step 3: Apply the integration by parts formula

Substitute into the integration by parts formula: \[ \int 2xe^{4x} \, dx = uv - \int v \, du \] This becomes: \[ = (2x) \left( \frac{1}{4}e^{4x} \right) - \int \left( \frac{1}{4}e^{4x} \right) (2) \, dx \] Simplify to get: \[ = \frac{1}{2}xe^{4x} - \frac{1}{2} \int e^{4x} \, dx \]

Step 4: Evaluate the remaining integral

Evaluate the integral \( -\frac{1}{2} \int e^{4x} \, dx \). This integral is \( \int e^{4x} \, dx = \frac{1}{4}e^{4x} + C \), so the expression becomes \[ -\frac{1}{2} \cdot \frac{1}{4}e^{4x} = -\frac{1}{8} e^{4x} \]

Step 5: Write the final result

Combine the results to write the final expression for the integral: \[ \frac{1}{2}xe^{4x} - \frac{1}{8}e^{4x} + C \]

Step 6: Check by differentiating

Differentiate the result: \( \frac{d}{dx} \left( \frac{1}{2}xe^{4x} - \frac{1}{8}e^{4x} \right) \). Use the product rule and chain rule to show that differentiating gives back the original integrand \( 2xe^{4x} \).

Key Concepts

Integration Techniques

Integration by parts is a powerful method used to solve integrals where direct integration is not straightforward. It is derived from the product rule for differentiation and allows us to transform a complicated integral into simpler parts. Here, to evaluate \( \int 2xe^{4x} \, dx \), we use integration by parts, which is presented by the formula:
\[ \int u \, dv = uv - \int v \, du \]
To apply this technique, we must first identify parts of the integrand that can be set as \( u \) and \( dv \). In our example, select \( u = 2x \) and \( dv = e^{4x} \, dx \).

• Differentiate \( u = 2x \) to get \( du = 2 \, dx \).
• Integrate \( dv = e^{4x} \, dx \) resulting in \( v = \frac{1}{4} e^{4x} \).

Once \( u \), \( dv \), \( du \), and \( v \) are determined, substitution into the integration by parts formula separates the integral into a more manageable expression that is then evaluated.

Differentiation

Differentiation is the process of finding the derivative of a function, which effectively measures how the function value changes as its input changes. In the context of integration by parts, differentiation is essential for computing the term \( du \).

To differentiate a function like \( u = 2x \), we apply basic differentiation rules. Since \( u = 2x \) is a simple linear function, its derivative is straightforward:
\[ du = 2 \, dx \]
This step is key in preparing to use the integration by parts formula as it provides the necessary component \( du \). Differentiation also plays a crucial role when it comes to verifying our integration work. By differentiating the integrated result, we should obtain the original integrand. For instance, if our integral solution was correctly found as \( \frac{1}{2}xe^{4x} - \frac{1}{8}e^{4x} + C \), differentiating this should yield the original function \( 2xe^{4x} \). This ensures the solution's accuracy.

Calculus Problem Solving

Calculus problem-solving is enhanced by employing a step-by-step methodology to approach complex integrals or derivatives. Through careful selection of techniques like integration by parts, problems that appear difficult become manageable.

In solving \( \int 2xe^{4x} \, dx \), initially identify suitable parts \( u \) and \( dv \) for the integration by parts framework. It's crucial to pick functions that simplify the remaining integral after applying the formula. The process unfolds over:

• Setting \( u = 2x \) simplifies \( du \) to \( 2 \, dx \).
• Choosing \( dv = e^{4x} \, dx \) allows a resulting \( v \) that aligns with the integration technique.
• Simplifying and solving each step progressively tests understanding of each calculus concept.

Verification, an important step, involves differentiating the final solution to ensure the original integrals match. This confirms the understanding and application of calculus principles effectively. By breaking down the problem and utilizing the fundamental calculus rules, students can convert a daunting calculus question into a sequence of simpler, manageable tasks.