Math Problem Statement

What is the equation by parts

Solution

The integration by parts formula is a powerful technique for solving integrals where the standard methods (like basic integration rules or substitution) are difficult to apply. It comes from the product rule of differentiation and is useful when the integrand is 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 you choose to differentiate.
  • dvdv is a function that you choose to integrate.

Steps to apply integration by parts:

  1. Identify parts of the integrand as uu and dvdv.
  2. Differentiate uu to get dudu.
  3. Integrate dvdv to get vv.
  4. Substitute into the formula: udv=uvvdu\int u \, dv = uv - \int v \, du.
  5. Solve the resulting integral.

Example:

For the integral xexdx\int x e^x \, dx:

  • Let u=xu = x and dv=exdxdv = e^x \, dx.
  • Differentiate uu: du=dxdu = dx.
  • Integrate dvdv: v=exv = e^x.

Now apply the formula:

xexdx=xexexdx\int x e^x \, dx = x e^x - \int e^x \, dx

The remaining integral is easy to solve:

exdx=ex\int e^x \, dx = e^x

So the final solution is:

xexex+Cx e^x - e^x + C

Where CC is the constant of integration.

Would you like more details on how to apply this method or see another example?

5 related questions:

  1. How do you choose uu and dvdv in integration by parts?
  2. Can integration by parts be used multiple times for complex integrals?
  3. What are some common integrals that can be solved using integration by parts?
  4. How does the formula for integration by parts derive from the product rule?
  5. What happens if the resulting integral after applying parts is the same as the original?

Tip:

When choosing uu in integration by parts, try the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) to prioritize which function to differentiate.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration

Formulas

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

Theorems

Product Rule of Differentiation

Suitable Grade Level

Undergraduate Calculus