General solution of integration by parts of int(x^n*e^x)

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SUMMARY

The discussion focuses on deriving a general formula for the integral of the function \( \int (x^n e^x) \, dx \) using integration by parts. The user attempts to apply the integration by parts formula \( \int(uv')dx = uv - \int(vu')dx \) but struggles with the \( uv \) component. The key insight shared is the recursive relationship \( I_n = x^n e^x - n I_{n-1} \), which simplifies the integration process by relating it to the integral of a lower power of \( x \). This recursive formula is essential for deriving the general solution.

PREREQUISITES
  • Understanding of integration by parts, specifically the formula \( \int(uv')dx = uv - \int(vu')dx \)
  • Familiarity with recursive functions and their applications in calculus
  • Knowledge of exponential functions and their properties
  • Basic algebraic manipulation skills for simplifying expressions
NEXT STEPS
  • Study the derivation of the recursive formula \( I_n = x^n e^x - n I_{n-1} \)
  • Practice integration by parts with various functions to gain proficiency
  • Explore the application of integration by parts in solving differential equations
  • Learn about the use of generating functions in calculus for solving integrals
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for effective methods to teach integration by parts and recursive formulas.

brandy
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Homework Statement


i have to create a general formula for integral of (x^n * e^x) dx
using whatever method i deem appropriate. (the only way i could think of is by parts)


Homework Equations


int(x^n * e^x)dx
int(uv')dx=uv-int(vu')dx


The Attempt at a Solution


i used integration by parts. so. I am having trouble with the uv part.
so far I've got
n!*e^x * (U) - int(e^x*n!)
U=? something that sums up u-u'-u''-u'''... until x is to the power of 1.

i figured out the function n-(n-1)-(n-2) etc which is = n-n(n-1)/2
i think if i can manipulate it enough it can give me the solution. but idk how

really, i just need a push in the right direction. or some clues or hints or something. ps make it simple, i take a while to understand other peoples working.
 
Last edited:
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ooooh poooooo!
i just realized this doesn't factor in the fact that you end up minusing a new function with a negative in it and so on so the solution ends up as +term -term + term - term etc


also, something that just further comfused me was this:
i read the next part of the question which says that i need to derive this formula which is the solution to the problem


In = xn ex - n*In-1
where In [tex]\int[/tex]xn ex dx
 
brandy said:
In = xn ex - n*In-1
where In [tex]\int[/tex]xn ex dx

Just apply integration by parts once to In and that result would easily follow.
 

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