Math Problem Statement

Evaluate ∫(x^2 e^{-x}) dx using integration by parts method. Number and describe all 8-steps, as covered during lecture.

Solution

To evaluate the integral

x2exdx\int x^2 e^{-x} \, dx

using the integration by parts method, I'll outline the solution in 8 steps as requested.


Steps for Solution

  1. Identify uu and dvdv:

    • We choose u=x2u = x^2 and dv=exdxdv = e^{-x} \, dx.
    • This choice is based on the LIATE rule (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) which helps us decide the order for uu and dvdv.
  2. Differentiate uu and Integrate dvdv:

    • Differentiate uu to find dudu:
      u=x2du=2xdxu = x^2 \Rightarrow du = 2x \, dx
    • Integrate dvdv to find vv:
      dv=exdxv=exdv = e^{-x} \, dx \Rightarrow v = -e^{-x}
  3. Apply Integration by Parts Formula:

    • The formula for integration by parts is udv=uvvdu\int u \, dv = uv - \int v \, du.
    • Substitute uu, vv, dvdv, and dudu: x2exdx=x2ex(ex)(2x)dx\int x^2 e^{-x} \, dx = -x^2 e^{-x} - \int (-e^{-x}) (2x) \, dx
    • Simplify the expression: =x2ex+2xexdx= -x^2 e^{-x} + 2 \int x e^{-x} \, dx
  4. Set Up Second Integration by Parts:

    • For xexdx\int x e^{-x} \, dx, we apply integration by parts again.
    • Choose u=xu = x and dv=exdxdv = e^{-x} \, dx.
  5. Differentiate u=xu = x and Integrate dv=exdxdv = e^{-x} \, dx:

    • Differentiate uu:
      u=xdu=dxu = x \Rightarrow du = dx
    • Integrate dvdv:
      dv=exdxv=exdv = e^{-x} \, dx \Rightarrow v = -e^{-x}
  6. Apply Integration by Parts Again:

    • Substitute into the integration by parts formula: xexdx=xex(ex)dx\int x e^{-x} \, dx = -x e^{-x} - \int (-e^{-x}) \, dx
    • Simplify: =xex+exdx= -x e^{-x} + \int e^{-x} \, dx
  7. Evaluate the Remaining Integral:

    • Solve exdx\int e^{-x} \, dx: exdx=ex\int e^{-x} \, dx = -e^{-x}
    • Substitute back: xexdx=xexex=(x+1)ex\int x e^{-x} \, dx = -x e^{-x} - e^{-x} = -(x + 1)e^{-x}
  8. Combine All Parts:

    • Substitute back into the original integral: x2exdx=x2ex+2((x+1)ex)\int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2(-(x + 1)e^{-x})
    • Simplify: =x2ex2(x+1)ex= -x^2 e^{-x} - 2(x + 1)e^{-x}
    • Factor out exe^{-x}: =(x2+2x+2)ex= -(x^2 + 2x + 2)e^{-x}

Final Answer

x2exdx=(x2+2x+2)ex+C\int x^2 e^{-x} \, dx = -(x^2 + 2x + 2)e^{-x} + C

Would you like more details on any step?


Related Questions

  1. How does the LIATE rule help in choosing uu and dvdv in integration by parts?
  2. What are some alternative methods to evaluate integrals involving polynomials and exponentials?
  3. How would the process change if the integrand was x3exx^3 e^{-x} instead?
  4. Why is integration by parts repeated when the uu-term is still a polynomial?
  5. How can you check your answer after performing integration by parts?

Tip

For integration by parts, choose uu such that it becomes simpler when differentiated, allowing dvdv to be easily integrated.

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Math Problem Analysis

Mathematical Concepts

Integration by Parts
Exponential Functions
Polynomials

Formulas

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

Theorems

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Suitable Grade Level

College level calculus