Math Problem Statement

integration by parts

Integration by parts

Solution

The expression given in the image is:

(p44p312p224p24)ep+C(-p^4 - 4p^3 - 12p^2 - 24p - 24)e^{-p} + C

To determine if this expression involves integration by parts, we can break it down:

  1. Integration by parts is typically used when integrating a product of two functions. The formula for integration by parts is:

udv=uvvdu\int u \, dv = uv - \int v \, du

Where:

  • uu is a function that we differentiate,
  • dvdv is a function that we integrate.

The expression above has a polynomial multiplied by an exponential function, epe^{-p}, which often suggests the need for integration by parts, because integrating products like pnepp^n e^{-p} frequently benefits from this technique.

Let's reconstruct the process leading to this result:

  1. Initial form: The original integral was likely of the form:

(p4+4p3+12p2+24p+24)epdp\int (p^4 + 4p^3 + 12p^2 + 24p + 24) e^{-p} \, dp

  1. Application of Integration by Parts: This could be integrated step by step, using pnepp^n e^{-p} in each term, where u=pnu = p^n and dv=epdpdv = e^{-p} \, dp.

  2. Final Answer: The result of the integration is:

(p44p312p224p24)ep+C(-p^4 - 4p^3 - 12p^2 - 24p - 24)e^{-p} + C

This indicates a repeated application of integration by parts, where each polynomial term pnp^n gets reduced by differentiating it until the integral becomes easier to handle.

Would you like a detailed step-by-step breakdown of how to perform this integration using the method of parts?

Related Questions:

  1. How do you choose uu and dvdv in integration by parts?
  2. What happens if you repeatedly apply integration by parts on polynomials?
  3. Why is integration by parts particularly useful for polynomials times exponentials?
  4. Can integration by parts be automated for higher-order polynomials?
  5. How do you deal with constants like CC in indefinite integrals?

Tip:

Always try to choose uu as the function that simplifies when differentiated, and dvdv as the function that is easy to integrate!

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration by Parts
Polynomial Integration
Exponential Functions

Formulas

Integration by parts formula: ∫u dv = uv - ∫v du

Theorems

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Suitable Grade Level

Undergraduate Calculus