Examen de cálculo integral

instituto tecnológico superior de san Martin Texmelucan

ingeniería industrial

Alejandro cerón Ochoa

examen de cálculo integral

[pic 3]

Problema 57

[pic 4](x + 1)2 ex dx

Integración por partes [pic 5]uv`=uv -[pic 6]u`v        

u= (x+1)2        u`=2(x+1)                v`=ex                v=ex

= (x+1)2 ex -[pic 7]2(x+1) ex dx

[pic 8]2(x+1) ex dx  

Se saca la constante

2[pic 9](x+1) ex dx

Integración por partes  [pic 10]uv`=uv -[pic 11]u`v

u= (x+1),        u`=1                v`=ex        v=ex

=2 ( ( x+1) ex -  [pic 12]1 ex dx )

= 2 ( ex ( x + 1 ) - [pic 13]ex dx

Aplicando la regla de integración  [pic 14]ex dx = ex

=2 (ex ( x+1)-ex)

= (x+1)2 ex – 2 ( ex (x+1) – ex ) +C

Problema 56

[pic 15]x ln ( [pic 16]) dx    = [pic 17][pic 18] x ln ( x + 2 ) dx

Se saca la constante

[pic 19][pic 20]x ln (x+2) dx

Integración por partes  [pic 21]uv`=uv -[pic 22]u`v[pic 23]

u= ln (x+2)        u`= [pic 24]                v`=x                v= [pic 25][pic 26][pic 27]

= [pic 28] ( ln (x+2) [pic 29]  -  [pic 30][pic 31]  [pic 32]  dx)[pic 33]

= [pic 34] ([pic 35]x2 ln (x+2)  -  [pic 36][pic 37] dx

Sacar la constante[pic 38]

[pic 39] ([pic 40]x2 ln (x+2)  -  [pic 41][pic 42][pic 43] dx

Aplicar integración por sustitución  [pic 44] f (g(x) ). G`=(x) dx = [pic 45]f (u) du ,                u= g(x)

u= x+2                du= 1 dx                 dx= 1 du[pic 46]

=[pic 47]( [pic 48] x2 ln (x+2) - [pic 49] [pic 50][pic 51]   1dx )[pic 52]

=[pic 53]( [pic 54] x2 ln (x+2) - [pic 55] [pic 56][pic 57]   du )

u= x+2         x= u-2[pic 58][pic 59]

=[pic 60]( [pic 61] x2 ln (x+2) - [pic 62] [pic 63][pic 64]  du)

Simplificando

=[pic 65]( [pic 66] x2 ln (x+2) - [pic 67] [pic 68]u+[pic 69]-4du)

Aplicar la regla de la suma   [pic 70]f (x) ± g (dx= [pic 71]f (x) dx ± [pic 72]g (x) dx

=[pic 73]( [pic 74] x2 ln (x+2) - [pic 75] ([pic 76]u du +[pic 77][pic 78]du -[pic 79]4du) )

[pic 80]u du[pic 81]

Aplicando regla de la potencia [pic 82]xa dx =[pic 83]                a≠ -1                [pic 84]

= [pic 85][pic 86][pic 87]

= [pic 88]        

[pic 89][pic 90]du

Se saca la constante

4[pic 91][pic 92] du

Aplicando la regla de integración  [pic 93][pic 94]du= ln (u)[pic 95]

= 4 ln (u)

[pic 96]4du                        [pic 97]f (a) dx= x. f (a)[pic 98]

= 4u

[pic 99]

=[pic 100]( [pic 101] x2 ln (x+2) - [pic 102] ([pic 103]+ 4 ln (u) – 4 u) ) 

Sustituimos u= x+2[pic 104]

=[pic 105]( [pic 106] x2 ln (x+2) - [pic 107] (([pic 108])+ 4 ln (x+2) - 4 (x+2) -))

Simplificamos [pic 109][pic 110]

=[pic 111]( [pic 112] x2 ln (x+2) + [pic 113] (- [pic 114] (x+2)+ 4 (x+2) - 4 ln (x+2) ))+C

Problema 66

[pic 115]

Integración por método de sustitución [pic 116] f (g(x) ). G`=(x) dx = [pic 117]f (u) du ,                u= g(x)

u=  [pic 118]:          du= [pic 119]             du=         [pic 120],                dx= u du

=[pic 121]xu udu

=[pic 122]xu2 du

u= [pic 123]         x=[pic 124](u2 – 1 )[pic 125]

= [pic 126][pic 127](u2 – 1 ) u2 du

Sacamos la constante

[pic 128][pic 129](u2 – 1) u2 du

=[pic 130][pic 131](u4 –u2 )du

Aplicando regla de la suma

=[pic 132]([pic 133]u4 du -  [pic 134]u2 du)

[pic 135][pic 136][pic 137]

[pic 138]u4 du = [pic 139]    =   [pic 140]

[pic 141][pic 142][pic 143]

[pic 144]u2 du = [pic 145]   =  [pic 146]

[pic 147][pic 148]

=[pic 149]([pic 150] -  [pic 151] )

Sustituir   u= [pic 152][pic 153][pic 154]

=[pic 155]( [pic 156]

Simplificar [pic 157]

= [pic 158]([pic 159][pic 160]    -  [pic 161][pic 162])+C

...