3ra de Metodos
1
: ñ . : 20071024 ∫ − () ∞ 0 − , ó () = − : () = − ( − ) . ó: = 2: 2 () = − 2 2 ( 2 − )= − [ − ( 2 − 4 + 2)] = −2 ( 2 − 4 + 2) ó: 2 − 4 + 2 = 0 , 1 = 0.58578644 , 2 = 3.41421356 : 1 = 0.55666790 , 2 = 0.03290227 : ∫ − () ∞ 0 = 0.55666790 × (0.58578644) + 0.03290227 × (3.41421356) ∫ − () ∞ 0 = 0.29889741 ≈ 0.2989 : = 0.5 − 0.2989 = 0.2011 ------------------------------------------------------------------------------------------------------------------------------ : () = ( − ) ; = (!) 2 [ ′ ()] 2 = 2: 2 () = 2 2 ( 2 − )= [ − ( 2 − 4 + 2)] = 2 − 4 + 2 = 4 (2 − 4) 2 ó: 2 − 4 + 2 = 0 , 1 = 0.58578644 , 2 = 3.41421356 : 1 = 0.85355339 , 2 = 0.14644661 ∫ − () ∞ 0 = 0.85355339 × (0.58578644) + 0.14644661 × (3.41421356) ∫ − () ∞ 0 = 0.43245945 ≈ 0.4325 : = 0.5 − 0.4325 = 0.0675
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Transcript of 3ra de Metodos
𝑁𝑜𝑚𝑏𝑟𝑒: 𝐴𝑙𝑒𝑥 𝑆𝑎𝑚𝑢𝑒𝑙 𝐿𝑢𝑑𝑒ñ𝑎 𝐻𝑢𝑎𝑚𝑎𝑛𝑖. 𝐶𝑜𝑑𝑖𝑔𝑜: 20071024𝐷
𝐴𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑟 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 ∫ 𝑒−𝑥𝑠𝑒𝑛(𝑥)𝑑𝑥∞
0
𝑢𝑠𝑎𝑛𝑑𝑜 𝑢𝑛𝑎 𝑐𝑢𝑎𝑑𝑟𝑎𝑡𝑢𝑟𝑎 𝑑𝑒 𝐺𝑎𝑢𝑠𝑠 −
𝐿𝑎𝑔𝑢𝑒𝑟𝑟𝑒 𝑑𝑒 𝑑𝑜𝑠 𝑝𝑢𝑛𝑡𝑜𝑠, 𝑑𝑜𝑛𝑑𝑒 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑑𝑒 𝑝𝑒𝑠𝑜 𝑒𝑠 𝜔(𝑥) = 𝑒−𝑥 𝑦 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑑𝑒
𝐿𝑎𝑔𝑢𝑒𝑟𝑟𝑒 𝑑𝑒 𝑔𝑟𝑎𝑑𝑜 𝑛 𝑣𝑖𝑒𝑛𝑒 𝑑𝑎𝑑𝑜 𝑝𝑜𝑟:
𝐿𝑛(𝑥) = 𝑒−𝑥𝑑𝑛
𝑑𝑥𝑛(𝑥𝑛𝑒−𝑥)
𝐶𝑎𝑙𝑐𝑢𝑙𝑒 𝑒𝑙 𝑒𝑟𝑟𝑜𝑟 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜 𝑐𝑜𝑛 𝑟𝑒𝑑𝑜𝑛𝑑𝑒𝑜 𝑎𝑙 𝑐𝑢𝑎𝑟𝑡𝑜 𝑑𝑒𝑐𝑖𝑚𝑎𝑙.
𝑆𝑜𝑙𝑢𝑐𝑖ó𝑛:
𝑃𝑎𝑟𝑎 𝑑𝑜𝑠 𝑝𝑢𝑛𝑡𝑜𝑠 𝑛 = 2:
𝐿2(𝑥) = 𝑒−𝑥𝑑2
𝑑𝑥2(𝑥2𝑒−𝑥) = 𝑒−𝑥[𝑒−𝑥(𝑥2 − 4𝑥 + 2)] = 𝑒−2𝑥(𝑥2 − 4𝑥 + 2)
𝑅𝑒𝑠𝑜𝑙𝑣𝑖𝑒𝑛𝑑𝑜 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛: 𝑥2 − 4𝑥 + 2 = 0 , 𝑥1 = 0.58578644 , 𝑥2 = 3.41421356
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑙𝑜𝑠 𝑝𝑒𝑠𝑜𝑠: 𝜔1 = 0.55666790 , 𝜔2 = 0.03290227
𝐻𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑙𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙:
∫ 𝑒−𝑥𝑠𝑒𝑛(𝑥)𝑑𝑥∞
0
= 0.55666790 × 𝑠𝑒𝑛(0.58578644) + 0.03290227 × 𝑠𝑒𝑛(3.41421356)
∫ 𝑒−𝑥𝑠𝑒𝑛(𝑥)𝑑𝑥∞
0
= 0.29889741 ≈ 0.2989
𝐻𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑒𝑙 𝑒𝑟𝑟𝑜𝑟 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜: 𝐸 = 0.5 − 0.2989 = 0.2011
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𝑅𝑒𝑠𝑜𝑙𝑣𝑖𝑒𝑛𝑑𝑜 𝑒𝑙 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑛𝑑𝑜:
𝐿𝑛(𝑥) = 𝑒𝑥𝑑𝑛
𝑑𝑥𝑛(𝑥𝑛𝑒−𝑥) ; 𝜔 =
(𝑛!)2
𝑥[𝐿𝑛′ (𝑥)]2
𝑃𝑎𝑟𝑎 𝑑𝑜𝑠 𝑝𝑢𝑛𝑡𝑜𝑠 𝑛 = 2:
𝐿2(𝑥) = 𝑒𝑥𝑑2
𝑑𝑥2(𝑥2𝑒−𝑥) = 𝑒𝑥[𝑒−𝑥(𝑥2 − 4𝑥 + 2)] = 𝑥2 − 4𝑥 + 2
𝜔 =4
𝑥(2𝑥 − 4)2
𝑅𝑒𝑠𝑜𝑙𝑣𝑖𝑒𝑛𝑑𝑜 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛: 𝑥2 − 4𝑥 + 2 = 0 , 𝑥1 = 0.58578644 , 𝑥2 = 3.41421356
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑙𝑜𝑠 𝑝𝑒𝑠𝑜𝑠: 𝜔1 = 0.85355339 , 𝜔2 = 0.14644661
∫ 𝑒−𝑥𝑠𝑒𝑛(𝑥)𝑑𝑥∞
0
= 0.85355339 × 𝑠𝑒𝑛(0.58578644) + 0.14644661 × 𝑠𝑒𝑛(3.41421356)
∫ 𝑒−𝑥𝑠𝑒𝑛(𝑥)𝑑𝑥∞
0
= 0.43245945 ≈ 0.4325
𝐻𝑎𝑙𝑙𝑎𝑛𝑑𝑜 𝑒𝑙 𝑒𝑟𝑟𝑜𝑟 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜: 𝐸 = 0.5 − 0.4325 = 0.0675
