Embed Size (px)
DESCRIPTION
.
Citation preview
𝐼 = ∫√2 − 𝑥 − 𝑥2
𝑥2𝑑𝑥
𝐶𝑜𝑚𝑝𝑙𝑒𝑡𝑎𝑛𝑑𝑜 𝑐𝑢𝑎𝑑𝑟𝑎𝑑𝑜𝑠: 2 − 𝑥 − 𝑥2 = −(𝑥2 + 𝑥 − 2) = − [(𝑥2 + 2(1
2)𝑥 +
1
4) −
1
4− 2]
= −[(𝑥 +1
2)2
−9
4] =
9
4− (𝑥 +
1
2)2
𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑦𝑒𝑛𝑑𝑜 𝑒𝑛 𝑙𝑎 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙
𝐼 = ∫√9
4− (𝑥 +
12)2
𝑥2𝑑𝑥
𝐻𝑎𝑐𝑖𝑒𝑛𝑑𝑜 𝑒𝑙 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 1: 𝑦 = 𝑥 +1
2 ; 𝑑𝑦 = 𝑑𝑥 ; 𝑑𝑒 𝑑𝑜𝑛𝑑𝑒 𝑥 = 𝑦 −
1
2
𝐼 = ∫√9
4− 𝑦2
(𝑦 −12)2 𝑑𝑦
𝐴𝑝𝑙𝑖𝑐𝑎𝑚𝑜𝑠 𝑆𝑢𝑠𝑡𝑖𝑢𝑐𝑖𝑜𝑛 𝑡𝑟𝑖𝑔𝑜𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐𝑎: 𝑦 =3
2𝑠𝑒𝑛𝜃 ; 𝑑𝑦 =
3
2𝑐𝑜𝑠𝜃𝑑𝜃 ;=>
9
4− 𝑦2 =
9
4𝑐𝑜𝑠2𝜃
𝐼 = ∫9𝑐𝑜𝑠2𝜃
4 (32 𝑠𝑒𝑛𝜃 −
12)
2 𝑑𝜃 =9
4∫
𝑐𝑜𝑠2𝜃
(3𝑠𝑒𝑛𝜃 − 1)2
4
𝑑𝜃 = 9∫𝑐𝑜𝑠2𝜃
(3𝑠𝑒𝑛𝜃 − 1)2𝑑𝜃
𝐴ℎ𝑜𝑟𝑎 𝑡𝑟𝑎𝑏𝑎𝑗𝑎𝑟𝑒𝑚𝑜𝑠 𝑝𝑜𝑟 𝑝𝑎𝑟𝑡𝑒𝑠
𝑠𝑒𝑎 𝑢 = 𝑐𝑜𝑠𝜃 ; 𝑑𝑢 = −𝑠𝑒𝑛𝜃𝑑𝜃 ; 𝑑𝑣 =𝑐𝑜𝑠𝜃
(3𝑠𝑒𝑛𝜃 − 1)2𝑑𝜃
𝑣 = ∫𝑐𝑜𝑠𝜃
(3𝑠𝑒𝑛𝜃 − 1)2𝑑𝜃 = −
1
3(3𝑠𝑒𝑛𝜃 − 1)+ 𝑐
𝐼 = 9∫𝑐𝑜𝑠2𝜃
(3𝑠𝑒𝑛𝜃 − 1)2𝑑𝜃 = −
𝑐𝑜𝑠𝜃
3(3𝑠𝑒𝑛𝜃 − 1)− ∫
𝑠𝑒𝑛𝜃
3(3𝑠𝑒𝑛𝜃 − 1)𝑑𝜃
𝑀𝑎𝑛𝑖𝑝𝑢𝑙𝑎𝑚𝑜𝑠 𝑢𝑛 𝑝𝑜𝑐𝑜 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙
𝐼 = 9∫𝑐𝑜𝑠2𝜃
(3𝑠𝑒𝑛𝜃 − 1)2𝑑𝜃 = −
𝑐𝑜𝑠𝜃
3(3𝑠𝑒𝑛𝜃 − 1)−
1
3[1
3∫
3𝑠𝑒𝑛𝜃 − 1 + 1
(3𝑠𝑒𝑛𝜃 − 1)𝑑𝜃]
𝐼 = 9∫𝑐𝑜𝑠2𝜃
(3𝑠𝑒𝑛𝜃 − 1)2𝑑𝜃 = −
𝑐𝑜𝑠𝜃
3(3𝑠𝑒𝑛𝜃 − 1)−
1
9[∫
3𝑠𝑒𝑛𝜃 − 1
(3𝑠𝑒𝑛𝜃 − 1)𝑑𝜃 + ∫
1
(3𝑠𝑒𝑛𝜃 − 1)𝑑𝜃]
𝐼 = 9∫𝑐𝑜𝑠2𝜃
(3𝑠𝑒𝑛𝜃 − 1)2𝑑𝜃 = −
𝑐𝑜𝑠𝜃
3(3𝑠𝑒𝑛𝜃 − 1)−
1
9[∫𝑑𝜃 + ∫
1
(3𝑠𝑒𝑛𝜃 − 1)𝑑𝜃]
𝐼 = ∫𝑐𝑜𝑠2𝜃
(3𝑠𝑒𝑛𝜃 − 1)2𝑑𝜃 = −
3𝑐𝑜𝑠𝜃
(3𝑠𝑒𝑛𝜃 − 1)− [𝜃 + ∫
1
(3𝑠𝑒𝑛𝜃 − 1)𝑑𝜃]
𝑆𝑒𝑎 𝐴 = ∫1
(3𝑠𝑒𝑛𝜃 − 1)𝑑𝜃
𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑟𝑒𝑚𝑜𝑠 𝑒𝑙 𝑚𝑒𝑡𝑜𝑑𝑜 𝑑𝑒 𝐶𝑎𝑚𝑏𝑖𝑜 𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙
𝑆𝑒𝑎 𝑡 = 𝑡𝑔 (𝜃
2) ; 𝑠𝑒𝑛𝜃 =
2𝑡
1 + 𝑡2 ; 𝑑𝜃 =
2
1 + 𝑡2𝑑𝑡
𝐴 = ∫1
(6𝑡
1 + 𝑡2 − 1)(
2
1 + 𝑡2)𝑑𝑡 = 2∫
1
(6𝑡 − 1 − 𝑡2)𝑑𝑡
𝐶𝑜𝑚𝑝𝑙𝑒𝑡𝑎𝑚𝑜𝑠 𝑐𝑢𝑎𝑑𝑟𝑎𝑑𝑜𝑠 𝑒𝑛 𝑒𝑙 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑑𝑜𝑟
(6𝑡 − 1 − 𝑡2) = −(𝑡2 − 6𝑡 + 1) = −[(𝑡2 − 2(3)𝑡 + 9) − 9 + 1] = −[(𝑡 − 3)2 − 8] = 8 − (𝑡 − 3)2
𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑖𝑚𝑜𝑠 𝑒𝑛 𝐴
𝐴 = 2∫1
8 − (𝑡 − 3)2𝑑𝑡
𝐻𝑎𝑐𝑒𝑚𝑜𝑠 𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑐𝑖𝑜𝑛 𝑇𝑟𝑖𝑔𝑜𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐𝑎: (𝑡 − 3) = √8𝑠𝑒𝑛𝛿 ; 𝑑𝑡 = √8𝑐𝑜𝑠𝛿𝑑𝛿 ; 8 − (𝑡 − 3)2 = 8𝑐𝑜𝑠2𝛿
𝐴 = 2∫√8𝑐𝑜𝑠𝛿𝑑𝛿
8𝑐𝑜𝑠2𝛿=
√8
4∫𝑠𝑒𝑐𝛿𝑑𝛿 =
√8
4𝐿𝑛|𝑠𝑒𝑐𝛿 + 𝑡𝑔𝛿| + 𝐶
𝑠𝑒𝑛𝛿 =𝑡 − 3
√8 ; 𝑠𝑒𝑐𝛿 =
√8
√8 − (𝑡 − 3)2 ; 𝑡𝑔𝛿 =
𝑡 − 3
√8 − (𝑡 − 3)2
𝐴 =√8
4𝐿𝑛 |
√8
√8 − (𝑡 − 3)2+
𝑡 − 3
√8 − (𝑡 − 3)2| + 𝑐 =
√8
4𝐿𝑛 |
√8 + 𝑡 − 3
√6𝑡 − 1 − 𝑡2|
𝐷𝑒𝑣𝑜𝑙𝑣𝑖𝑒𝑛𝑑𝑜 𝑒𝑙 𝐶𝑎𝑚𝑏𝑖𝑜 𝑈𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙:
𝐴 =√8
4𝐿𝑛 ||
√8 + 𝑡𝑔 (𝜃2) − 3
√6𝑡𝑔 (𝜃2) − 1 − 𝑡𝑔2 (
𝜃2)
|| + 𝐶
𝐸𝑛𝑡𝑜𝑛𝑐𝑒𝑠:
𝐼 = −3𝑐𝑜𝑠𝜃
(3𝑠𝑒𝑛𝜃 − 1)−
[
𝜃 +√8
4𝐿𝑛 ||
√8 + 𝑡𝑔 (𝜃2) − 3
√6𝑡𝑔 (𝜃2) − 1 − 𝑡𝑔2 (
𝜃2)
||
]
+ 𝐶
𝐷𝑒𝑣𝑜𝑙𝑣𝑖𝑒𝑛𝑑𝑜 𝑒𝑙 1𝑒𝑟 𝐶𝑎𝑚𝑏𝑖𝑜 𝑇𝑟𝑖𝑔𝑜𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐𝑜:
𝑠𝑒𝑛𝜃 =𝑦
(32) ; 𝑐𝑜𝑠𝜃 =
2√94 − 𝑦2
3 ; 𝜃 = 𝐴𝑟𝑐𝑠𝑒𝑛 (
2𝑦
3)
𝐼 = −2√9
4 − 𝑦2
(2𝑦 − 1)−
[
𝐴𝑟𝑐𝑠𝑒𝑛 (2𝑦
3) +
√8
4𝐿𝑛
|
|
| √8 + 𝑡𝑔 (𝐴𝑟𝑐𝑠𝑒𝑛 (
2𝑦3
)
2 ) − 3
√6𝑡𝑔(𝐴𝑟𝑐𝑠𝑒𝑛 (
2𝑦3 )
2) − 1 − 𝑡𝑔2 (
𝐴𝑟𝑐𝑠𝑒𝑛 (2𝑦3 )
2)|
|
|
]
+ 𝑐
𝐷𝑒𝑣𝑜𝑙𝑣𝑖𝑒𝑛𝑑𝑜 𝑒𝑙 𝑐𝑎𝑚𝑏𝑖𝑜 1: 𝑦 = 𝑥 +1
2
𝐼 = −2√2 − 𝑥 − 𝑥2
2𝑥−
[
𝐴𝑟𝑐𝑠𝑒𝑛 (2𝑥 + 1
3) +
√8
4𝐿𝑛
|
|
| √8 + 𝑡𝑔 (𝐴𝑟𝑐𝑠𝑒𝑛 (
2𝑥 + 13 )
2 ) − 3
√6𝑡𝑔(𝐴𝑟𝑐𝑠𝑒𝑛 (
2𝑥 + 13 )
2 ) − 1 − 𝑡𝑔2 (𝐴𝑟𝑐𝑠𝑒𝑛 (
2𝑥 + 13 )
2 )|
|
|
]
+ 𝑐
𝑭𝑰𝑵𝑨𝑳𝑰𝒁𝑨𝑵𝑫𝑶:
𝐼 = ∫√2 − 𝑥 − 𝑥2
𝑥2𝑑𝑥 = −
√2 − 𝑥 − 𝑥2
𝑥− 𝐴𝑟𝑐𝑠𝑒𝑛 (
2𝑥 + 1
3) −
√8
4𝐿𝑛
|
|
| √8 + 𝑡𝑔 (𝐴𝑟𝑐𝑠𝑒𝑛 (
2𝑥 + 13
)
2) − 3
√6𝑡𝑔 (𝐴𝑟𝑐𝑠𝑒𝑛 (
2𝑥 + 13 )
2) − 1 − 𝑡𝑔2 (
𝐴𝑟𝑐𝑠𝑒𝑛 (2𝑥 + 1
3 )
2)|
|
|
+ 𝐶
