Post on 13-Jul-2015
Taller Calculo Integral Parte 2
1.
1 β12
3(6) [π(0) + 4(0,08) + 2π(0,16) + 4π(0,24) + 2π(0.32) + 4π(0,4) + π(0,48)]
1 β12
3(6)[(
2
1 + (0)4)+ 4 (
2
1 + (0,08)4) + 2 (
2
1 + (0,16)4)+ 4 (
2
1 + (0,24)4) + 2 (
2
1 + (0,32)4)
+ 4 (2
1 + (0,4)4)+ (
2
1 + (0,48)4)]
12
18[2 + 4 (
2
1,0256) + 2 (
2
1,0066) + 4 (
2
1,0033) + 2 (
2
1,010) + 4 (
2
1,0256) + (
2
1,0530)]
1
36[2 + 4(1,9500) + 2(1,9868) + 4(1,9934) + (1,9801) + 4(1,9500 + (18993)]
1
36[2 + 7,8 + 3,9 + 7,9 + 3,9 + 7,8 + 1,8993] = [35,1993] = 0,972
B)
π(π, β) =β
2[π(π) + π(π)]
πΏπ₯
2[π(π₯0 + 2π(π₯1) + 2π(π₯2)β¦ π(π₯πβ1) + π(π₯π)]
1
24
π(π₯0) =2
1 + 0= 2
π(π₯1) =2
1 + 0,08= 1,9500
π(π₯2) =2
1 + 0,16= 1,9969
π(π₯3) =2
1 + 0,24= 1,9934
π(0,32) =2
1 + 0,34= 1,9801
π(0,4) =2
1 + 0,4= 1,9501
π(0,48) =2
1 + 0,48= 1,8993
0,08
2[2 + 2(1,9500) + 2(1,9868) + 2(1,9934) + 2(1,9801) + 2(1,9501) + 1,8993]
[2 + 3,9 + 3,97 + 3,98 + 3,96 + 3,9 + 1,89]
1
24[23,6]π’2
0,9833
2.
π β ππ = π
ππ = π
π = Β±βπ
π = Β±π
ππ β ππ = π
(π + π)(π β π) = π
π + π = π βπ β π = π
π = βπ β π = π
β« π β πππ ππ
π=
π β« π ππ
πβ β« πππ π
π
π=
ππ βππ
π|π
π =
π βπ
πβ π = π. ππππππ
3.
A.
π¦ = π₯2 β«β«β« π¦ =2
π₯2 + 1
π₯2 =2
π₯2 + 1
π₯2(π₯2 + 1) = 2
π₯4 + π₯2 β 2 = 0
(π₯2 + 2)(π₯2 β 1) = 0
π₯2 + 2 = 0 β π₯2 β 1 = 0
π₯ = Β±β2 β π₯ = Β±β1
π = Β±π
β« (2
π₯2 + 1) β (π₯2)ππ₯
1
β1
β« (2
π₯2 + 1) ππ₯ β β« π₯2ππ₯
1
β1
1
β1
= 2 ln(π₯2 + 1) βπ₯3
3 | β1
1
[ 2 ln(12 + 1) β13
3 ]
β [2 ln(β12 + 1) ββ13
3 ]
1.052961 β 1.719627
= βπ. ππππ
B.
π = ππ + π β«β«β« π = πππ β«β«
β« π = π π β«β«β« π
= π
ππ + π = ππ + π
ππ + π β π β ππ = π
ππ β ππ β π = π
(π β π)(π + π) = π
π = π β π = βπ
β« [ (ππ + π) β (ππ + π)]π
π
β« (ππ + π β ππ β π )π ππ
π
β« (βππ + ππ + π)π
ππ π
βπ β« (ππ β ππ β π)π ππ
π
βπ [β« πππ π β β« ππ π π β β« ππ ππ
π
π
π
π
π
]
βπ [ππ
πβ πππ β ππ + πͺ] |π
π
= βπ [ππ
πβ π β ππ β π β π]
+ π [ππ
πβ π β ππ β π β π]
= βπ[ βππ ] + π[ βπ. ππππ ]
= ππ. πππππ
C.
ππ = π β π β«β«β« π = π β π
π = βπ β π
π β π = π β π
π β π β π + π = π
β«β« βπ β π = π β π
(βπ β π β π β π)π
= ππ
π β π β ππ + π = π
βππ + π + π = π
π₯ =βπ Β± βπ2 β 4ππ
2π
π = βπ Β±βπ β π(βπ)(π)
π(βπ)
=βπ
π ππππ(π½)π π½
D)
π¦ = πππ π₯ β«β«β« π¦ = π πππ₯ , π₯ β [0,2π]
β« (πππ π₯ β π πππ₯)ππ₯2π
0
= β« πππ π₯ππ₯ β2π
0 β« π πππ₯ππ₯2π
0
= π πππ₯ + πππ π₯|02
= π ππ(2π) + cos(2π) = 0.0985 + 0.9951 = 1.0936 unidades cuadradas
Cos x:
X -3 -2 -1 0 1 2 3
Y 0.9988 0.9995 0.9998 1 0.9998 0.9995 0.9988
Sen x:
X -3 -2 -1 0 1 2 3
y -0.04 -
0.03
-
0.01
0 0.01 0.03 0.04
5.
π¦ = 4π₯ β π₯2
π₯2 β 4π₯ = 0
π₯(π₯ β 4) = 0
π₯ = 0 βπ₯ = 4
β« 4π₯ β π₯24
0ππ₯
β« 4π₯ππ₯4
0β β« π₯2ππ₯
4
0
4π₯2 β π₯3
3 |0
4
64 β 64
3β 0 = ππ. πππππ