

Integración por partes
Presentation
•
Mathematics
•
8th Grade
•
Easy
Lucia Baez
Used 7+ times
FREE Resource
8 Slides • 6 Questions
1
Integración por partes

2
Open Ended
Si tenemos ∫x2cos(x)dx a quién sugerimos como u y quién como dv
3
Multiple Choice
Si para ∫x2cos(x)dx decimos que u= x2 y dv= cos (x)dx, entonces du y v serán...
du=2x dx, v= ∫cos(x)dx=sin(x)
du=3x3 dx , v= cos(x)
du= 2 dx, v= ∫sin(x)dx=−cos(x)
4
∫x2cos(x)dx
aplicando la fórmula ∫u⋅dv=u⋅v−∫vdu
5
∫x2cos(x)dx
∫u⋅dv=u⋅v−∫vdu
= x2 sin(x)- ∫sin(x)2x dx
6
tenemos que: ∫2x sin(x) dx
La integral es más sencilla que la inicial, pero no es inmediata, por tanto debemos resolver por partes
7
Entonces: ∫2xsin(x)dx
por la propiedad de la integral 2∫xsin(x) dx
para esta integral u=x por lo que du= dx
dv= sin(x)dx por lo que v= ∫sin(x) dx=−cos(x)
8
lo anterior sustituido en ∫udv=uv−∫vdu
= -xcos(x)- ∫−cos(x)dx
=-xcos(x)+sin (x)
entonces, sustituimos este último resultado en la integral
9
x2sin(x)−∫2xsin(x)dx=
x2sin(x)−2(xcos(x)+sin(x))+c
10
Open Ended
Si tenemos ∫x2exdx escribe u, du, dv, v
11
Open Ended
Si en la integral anterior, u=x2,du=2xdx, dv=ex, v=ex, sustituye en la fórmula de integración por partes
12
Open Ended
Notarás que ∫ex2x dx no es una integral inmediata, por lo que hay que resolver por partes, entonces escribe u, du, dv, v
13
Open Ended
Si definimos que u=2x, du=2dx, dv=exdx, v=ex sustituye en la fórmula de integración por partes
14
Entonces x2ex−∫ex2x dx =
x2ex−2xex−2ex+c
Integración por partes

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