so, did you use the table? It will probably provide a reduction formula, such as
∫x^n e^ax dx = 1/a x^n e^ax - n/a ∫ x^(n-1) e^ax dx
this is achieved using integration by parts. So, going through that three times gives you
(1/9 x^3 - 1/27 x^2 + 2/243 x - 2/2187) e^9x + C
Use the table of integrals to find:
∫ x³*e⁹ˣ dx
2 answers
I came across this young man with his remarkable youtube math movies
called "blackpenredpen".
He shows a short-cut method to Integration By Parts calling it the DI method,
also called the Tabular Method. He uses 3 different ways in which the process
will end.
It works really well here:
https://www.youtube.com/watch?v=2I-_SV8cwsw&t=695s
∫ x³*e⁹ˣ dx = (1/9)x^3 e^(9x) - (1/27)x^2 e^(9x) + (2/243)x e^(9x) - (6/6561) e^(9x)
= (1/9)e^(9x) ( x^3 - (1/3)x^2 + (2/27)x - 2/243) + c
---D ----- I
+ | x^3 -- e^(9x)
- | 3x^2 -- (1/9)e^(9x)
+ | 6x ----- (1/81)e^(9x)
- | 6 ------- (1/729)e^(9x)
+ | 0 ------ (1/6561)e^(9x) <----- stop when you reach a zero in the D column, multiply the diagonals.
called "blackpenredpen".
He shows a short-cut method to Integration By Parts calling it the DI method,
also called the Tabular Method. He uses 3 different ways in which the process
will end.
It works really well here:
https://www.youtube.com/watch?v=2I-_SV8cwsw&t=695s
∫ x³*e⁹ˣ dx = (1/9)x^3 e^(9x) - (1/27)x^2 e^(9x) + (2/243)x e^(9x) - (6/6561) e^(9x)
= (1/9)e^(9x) ( x^3 - (1/3)x^2 + (2/27)x - 2/243) + c
---D ----- I
+ | x^3 -- e^(9x)
- | 3x^2 -- (1/9)e^(9x)
+ | 6x ----- (1/81)e^(9x)
- | 6 ------- (1/729)e^(9x)
+ | 0 ------ (1/6561)e^(9x) <----- stop when you reach a zero in the D column, multiply the diagonals.
1 answer