ap calc bc skill builder topic 6.3a

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Mastering AP Calculus BC: Skill Builder Topic 6.3a – Integration by Parts

This article serves as a comprehensive guide to AP Calculus BC Skill Builder Topic 6.3a, focusing on integration by parts. We'll break down the concept, provide examples, and offer strategies to master this crucial integration technique. Understanding integration by parts is vital for success in AP Calculus BC and beyond.

What is Integration by Parts?

Integration by parts is a powerful technique used to integrate products of functions. It's derived from the product rule for differentiation. Remember the product rule:

d(uv) = u dv + v du

Integrating both sides, we get:

∫d(uv) = ∫u dv + ∫v du

This simplifies to the integration by parts formula:

∫u dv = uv - ∫v du

The key to successfully applying integration by parts lies in choosing the appropriate 'u' and 'dv'. A helpful mnemonic is LIATE:

• Logarithmic functions
• Inverse trigonometric functions
• Algebraic functions
• Trigonometric functions
• Exponential functions

Generally, choose the function higher on the LIATE list as 'u'. This prioritizes functions that simplify when differentiated. Let's explore this with some examples.

Example 1: A Simple Application

Let's integrate ∫x cos(x) dx.

1. Choose u and dv: Let u = x (algebraic) and dv = cos(x) dx.

2. Find du and v: Then du = dx and v = ∫cos(x) dx = sin(x).

3. Apply the formula: ∫x cos(x) dx = x sin(x) - ∫sin(x) dx.

4. Solve the remaining integral: ∫sin(x) dx = -cos(x) + C.

5. Final answer: ∫x cos(x) dx = x sin(x) + cos(x) + C.

Example 2: Integration by Parts Multiple Times

Sometimes, you need to apply integration by parts more than once. Consider ∫x²eˣ dx.

1. First Integration by Parts: Let u = x² and dv = eˣ dx. Then du = 2x dx and v = eˣ. This gives us: ∫x²eˣ dx = x²eˣ - ∫2xeˣ dx

2. Second Integration by Parts: Now we need to integrate ∫2xeˣ dx. Let u = 2x and dv = eˣ dx. Then du = 2 dx and v = eˣ. This gives us: ∫2xeˣ dx = 2xeˣ - ∫2eˣ dx = 2xeˣ - 2eˣ + C

3. Combine the results: Substitute this back into the first equation: ∫x²eˣ dx = x²eˣ - (2xeˣ - 2eˣ) + C = x²eˣ - 2xeˣ + 2eˣ + C

Example 3: Integrating Inverse Trigonometric Functions

Let's tackle ∫x arctan(x) dx.

1. Choose u and dv: Let u = arctan(x) and dv = x dx.

2. Find du and v: du = 1/(1+x²) dx and v = x²/2.

3. Apply the formula: ∫x arctan(x) dx = (x²/2)arctan(x) - ∫(x²/(2(1+x²))) dx

4. Solve the remaining integral: This requires polynomial long division or rewriting the fraction. The result is: ∫(x²/(2(1+x²))) dx = (x/2) - (1/2)arctan(x) + C

5. Combine results: ∫x arctan(x) dx = (x²/2)arctan(x) - (x/2) + (1/2)arctan(x) + C

Tips for Success:

• Practice makes perfect: The more you practice, the better you'll become at choosing u and dv.
• Check your work: Differentiate your answer to ensure you get back to the original integrand.
• Don't be afraid to use substitutions: Sometimes, a substitution can simplify the integral before applying integration by parts.
• Utilize online resources: Khan Academy, Paul's Online Math Notes, and other online resources provide valuable practice problems and explanations.

Mastering integration by parts requires consistent practice and a good understanding of the underlying principles. By following these strategies and practicing regularly, you'll be well-equipped to tackle even the most challenging integration by parts problems on the AP Calculus BC exam. Remember to review your notes and consult your textbook or teacher for further clarification if needed. Good luck!