Chegaralanmagan funksiyaning xosmas integrali. Xosmas integralning bosh qiymati
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Chegaralanmagan funksiyaning xosmas integrali. Xosmas integralning bosh qiymati. Takrorlash. 1. Aniqmas integral; 2. Aniq integral; 3. Xosmas integrallar. 1-misol. Ta’rif yordamida hisoblang. 0 < 𝑎 < 𝑏 ∫ 𝑥
𝜆 𝑏 𝑎 𝑑𝑥
𝑃: 𝑎 = 𝑥 0 < 𝑥 1
2
𝑛 = 𝑏 𝑥 𝑘 = 𝑥 0 + 𝑘𝑑, 𝑑 = 𝑏 − 𝑎 𝑛
𝑥 𝑘 = 𝑥 0 𝑞 𝑘 , 𝑥 𝑛 = 𝑥 0 𝑞 𝑛 , 𝑏 = 𝑎𝑞 𝑛 ,
𝑛 = 𝑏 𝑎 , 𝑞 = √ 𝑏 𝑎
𝑃 = {𝑎, 𝑎𝑞, 𝑎𝑞 2 , … , 𝑎𝑞
𝑛−1 , 𝑏}, 𝜉 𝑘 = 𝑥
𝑘 = 𝑎𝑞
𝑛 , 𝑓(𝑥) = 𝑥 𝜆 𝜎 𝑃 (𝑓) = ∑ 𝑓(𝜉 𝑘 )𝛥𝑥 𝑘 𝑛 𝑘=1 = ∑ 𝑎 𝜆 𝑞 𝜆𝑘 (𝑥 𝑘 − 𝑥 𝑘−1
) 𝑛 𝑘=1 = ∑ 𝑎 𝜆 𝑞 𝜆𝑘 (𝑎𝑞
𝑘 − 𝑎𝑞
𝑘−1 ) 𝑛 𝑘=1 = = 𝑎 𝜆
1 𝑞 ) ∑ 𝑞 𝜆𝑘 𝑞 𝑘 𝑛 𝑘=1
= 𝑎 𝜆+1
(1 − 1 𝑞 ) ∑(𝑞 𝜆+1
) 𝑘 𝑛 𝑘=1 = [
𝑏 1 = 𝑞 𝜆+1 𝑞~𝑞
𝜆+1 ] = = 𝑎 𝜆+1
(1 − 1 𝑞 ) 𝑞 𝜆+1 (1 − 𝑞 𝑛(𝜆+1)
) 1 − 𝑞
𝜆+1 = 𝑎
𝜆+1 (1 −
1 𝑞 ) 𝑞 𝜆+1
(1 − 𝑞 𝑛𝜆 𝑞 𝑛 ) 1 − 𝑞 𝜆+1 = = 𝑎 𝜆+1
(1 − ( 𝑎 𝑏 ) 1 𝑛 ) ( 𝑏 𝑎) 𝜆+1
𝑛 (1 − (
𝑏 𝑎) 𝜆 𝑏 𝑎 ) 1 − ( 𝑏 𝑎) 𝜆+1 𝑛 = [
1 𝑛 → 0, 𝑎 𝑥 ~1 + 𝑥 ln 𝑎 𝑥 1 𝑛 ~1 +
1 𝑛 ln 𝑥 ( 𝑎 𝑏 ) 1 𝑛 ~1 + 1 𝑛 ln 𝑎 𝑏 ]
~ ~𝑎 𝜆+1
(− 1 𝑛 ln 𝑎 𝑏 ) (1 +
𝜆 + 1 𝑛 ln 𝑏 𝑎) (
1 − ( 𝑏 𝑎) 𝜆+1 ) − 𝜆 + 1 𝑛 ln 𝑏 𝑎 = [ln 𝑥 = ln 1 𝑥 ] → [𝑛 → ∞] → − 𝑎 𝜆+1 (1 − ( 𝑏 𝑎) 𝜆+1 ) 𝜆 + 1 = − 𝑎 𝜆+1 − 𝑏 𝜆+1
𝜆 + 1 = 𝑏 𝜆+1 − 𝑎
𝜆+1 𝜆 + 1
∫ 𝑥
𝜆 𝑏 𝑎 𝑑𝑥 = lim 𝜆 𝑃 →0 ∑ 𝑓(𝜉
𝑘 )𝛥𝑥
𝑘 𝑛 𝑘=1 = lim 𝑛→∞
∑ 𝑎 𝜆 𝑞 𝜆𝑘 (𝑎𝑞
𝑘 − 𝑎𝑞
𝑘−1 ) 𝑛 𝑘=1 = = lim 𝑛→∞
𝑎 𝜆+1
(1 − ( 𝑎 𝑏 ) 1 𝑛 ) ( 𝑏 𝑎) 𝜆+1
𝑛 (1 − (
𝑏 𝑎) 𝜆 𝑏 𝑎 ) 1 − ( 𝑏 𝑎) 𝜆+1 𝑛 = 𝑏 𝜆+1
− 𝑎 𝜆+1
𝜆 + 1
Javob: ∫ 𝑥 𝜆 𝑏 𝑎 𝑑𝑥 =
𝑏 𝜆+1
−𝑎 𝜆+1
𝜆+1
𝑓 ∈ 𝑅[𝑎, 𝑏] 𝑚 ≤ 𝑓(𝑥) ≤ 𝑀 , ∃𝜇 ∈ [𝑚, 𝑀] ∫ 𝑓(𝑥)𝑑𝑥 𝑏 𝑎 = 𝜇(𝑏 − 𝑎) 𝑓 ∈ 𝐶[𝑎, 𝑏] , ∃𝑐 ∈ [𝑚, 𝑀] ∫ 𝑓(𝑥)𝑑𝑥 𝑏 𝑎 = 𝑓(𝑐)(𝑏 − 𝑎)
𝑓, 𝑔 ∈ 𝑅[𝑎, 𝑏] , 0 < 𝑚 ≤ 𝑓(𝑥) ≤ 𝑀 , ∃𝜇 ∈ [𝑚, 𝑀] ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 𝑏 𝑎
𝑏 𝑎
𝑓 ∈ 𝐶[𝑎, 𝑏], 𝑔 ∈ 𝑅[𝑎, 𝑏] , 0 < 𝑓(𝑥) , ∃𝑐 ∈ [𝑎, 𝑏] ∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 𝑏 𝑎
𝑏 𝑎
𝜇 ∫ 𝑔(𝑥)𝑑𝑥 𝑏 𝑎
𝑏 𝑎 ≤ 𝛭 ∫ 𝑔(𝑥)𝑑𝑥 𝑏 𝑎
2-misol. ∫ 𝑒 𝑥 2 1 + 𝑥 2 1 𝑑𝑥 = ∫ 𝑒 𝑥 2 1 1 + 𝑥 𝑑𝑥 2 1 = [∃𝑐 ∈ [1; 2]] = 𝑒 𝑐 2
𝑑𝑥 1 + 𝑥
2 1 = 𝑒 𝑐 2 (ln 3 − ln 2) = = 𝑒 𝑐 2 ln 3 2 , 𝑒 𝑐 2 ∈ [𝑒, 𝑒
2 ] , 𝑒 ≤ 𝑒 𝑐 2
4
𝑒 ln 3 2
𝑒 𝑥
1 + 𝑥 2 1 𝑑𝑥 < 𝑒 4 ln 3 2
3-misol. 𝑎 ≤ 𝑡 ≤ 𝑏 𝑉 = ∫ 𝑆(𝑡)𝑑𝑡 𝑏 𝑎 3-misol. Quyidagi sirtlar bilan chegaralangan shaklni xajmini toping.
2 9 + 𝑦 2 4 − 𝑧 2 36 = −1,
𝑧 = 12 𝑧 = 𝑡
𝑥 2 9 + 𝑦 2 4 = 𝑡 2 36 − 1 ≥ 0 − ellips 𝑥 2 𝑎 2 + 𝑦 2 𝑏 2 = 1, 𝑆 = 𝜋𝑎𝑏 𝑥 2 (3√ 𝑡 2 36 − 1)
2 + 𝑦 2 (2√
𝑡 2 36 − 1) 2 = 1 𝑆(𝑡) = 𝜋 ∙ 3√ 𝑡 2 36 − 1 ∙ 2√ 𝑡 2
− 1 = 6𝜋 ( 𝑡 2 36 − 1)
𝑡 2 36 − 1 ≥ 0 , 𝑡 2 ≥ 36, 𝑡 ≥ 6 𝑉 = ∫ 6𝜋 ( 𝑡 2 36 − 1) 𝑑𝑡
12 6 = 6𝜋 ( 𝑡 3 108 − 𝑡)| 6 12 = 6𝜋 ∙ 8 = 48𝜋 4-misol. Xosmas integralni hisoblang: ∫ 𝑥 𝑒
(1 + 𝑥 2 ) 3 2 +∞ 1 𝑑𝑥 = ∫
𝑥 √(1 + 𝑥
2 ) 3 𝑒 arctan 𝑥
+∞ 1 𝑑𝑥 ∫ 𝑥 √(1 + 𝑥 2 ) 3 𝑒 arctan 𝑥
𝑑𝑥 =? 𝑥 √(1 + 𝑥 2 ) 3 𝑑𝑥 = 1 2 𝑑(1 + 𝑥 2 ) (1 + 𝑥 2 ) 3 2 = 1 2 (1 + 𝑥 2 ) − 3 2 𝑑(1 + 𝑥 2 ) = [1 + 𝑥 2 = 𝑡]
= 1 2 𝑡 − 3 2 𝑑𝑡 =
1 2 𝑑 ( 𝑡 − 1 2 − 1 2 ) = −𝑑𝑡
− 1 2 = −𝑑 1 √1 + 𝑥 2
𝑥 = tan 𝑡 ∫ 𝑒
arctan 𝑥
𝑑 1 √1 + 𝑥 2 = [1 + 𝑥
2 = 1 + tan 2 𝑡 =
1 cos
2 𝑡 , arctan 𝑥 = 𝑡] = = ∫ 𝑒 𝑡 𝑑 cos 𝑡 = [ 𝑢 = 𝑒 𝑡 𝑣 = cos 𝑡 𝑑𝑣 = 𝑒 𝑡 𝑑𝑡 ] = ⋯ = 𝑒 𝑡 2 (cos 𝑡 − sin 𝑡) + 𝐶 = = 𝑒
2 ( 1 √1 + 𝑥 2 − 𝑥 √1 + 𝑥
2 ) + 𝐶
∫ 𝑥 𝑒 arctan 𝑥 (1 + 𝑥 2 ) 3 2 +∞ 1 𝑑𝑥 = lim
𝐴→∞ ∫ 𝑥
𝑒 arctan 𝑥
(1 + 𝑥 2 ) 3 2 𝑑𝑥 𝐴 1 = lim 𝐴→∞ 𝑒 arctan 𝑡 2 ( 1 √1 + 𝑥 2 − 𝑥 √1 + 𝑥
2 )| 1 𝐴 =
= lim 𝐴→∞
𝑒 arctan 𝐴
2 ( 1 √1 + 𝐴 2 − 𝐴 √1 + 𝐴
2 ) = lim
𝐴→∞ 𝑒 arctan 𝐴 2 (
1 √1 + 𝐴
2 − 1 √ 1 𝐴 2 + 1) =
= 𝑒 𝜋 2 2 (0 − 1) = − 𝑒 𝜋 2 2
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