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Formulario cristian rmz
1.
C Calculo diferencial¡ FUETES: NOMBRE:
CRISTIAN JESUS RAMIREZ ALBA LIC. EDGAR MATA UNIVERSIDAD TECNOLOGICA DE TORREON
2.
PROBLEMARIO FORMULA 1 ∫ 𝐷𝑋
= 𝑋 + 𝐶 20𝐷𝑋 = 20 × +𝐶 17𝐷𝑋 = 17 × +𝐶 9𝐷𝑋 = 9𝑋 + 𝐶 50𝐷𝑋 = 50𝑋 + 𝐶 4𝐷𝑋 = 4𝑋 + 𝐶 FORMULA 2 ∫ 56 𝑋3 𝐷𝑋 = = 56 ∫ 𝑋3 𝐷𝑋 = 56 𝑋 4 4 + 𝐶 =14𝑋4 + 𝐶 ∫ 20 𝑋3 𝐷𝑋 = = 20 ∫ 𝑋3 𝐷𝑋 = 24 𝑋 4 4 + 𝐶 =5𝑋4 + 𝐶 ∫ 6 𝑋2 𝐷𝑋 = = 6 ∫ 𝑋2 𝐷𝑋 = 6 𝑋 3 3 + 𝐶 =2𝑋3 + 𝐶
3.
∫ 18 𝑋1 𝐷𝑋
= = 8 ∫ 𝑋1 𝐷𝑋 = 8 𝑋 2 2 + 𝐶 =4𝑋2 + 𝐶 ∫ 36 𝑋5 𝐷𝑋 = = 36 ∫ 𝑋5 𝐷𝑋 = 36 𝑋 6 6 + 𝐶 =6𝑋6 + 𝐶 FORMULA 3 ∫(𝐷𝑈 + 𝐷𝑉 + 𝐷𝑊) = ∫ 𝐷𝑈 + ∫ 𝐷𝑉 − ∫ 𝐷𝑊 ∫ (2+3𝑥2 − 8𝑥3 )𝑑𝑥 = ∫ 2𝑑𝑥 + ∫ 3𝑥2 𝑑𝑥 + ∫ − 8𝑥3 = 2𝑥 + 𝑐 + ∫ 3𝑥2 𝑑𝑥 ∫ − 8𝑥3 𝑑𝑥 = 2𝑥 + 𝑐 + 3 ∫ 𝑥2 𝑑𝑥 + ∫ − 8𝑥3 𝑑𝑥 = 2𝑥 + 𝑐 + 3 ( 1 3 𝑥3 + 𝑐) + ∫ − 8𝑥3 𝑑𝑥 = 2𝑥 + 𝑐 + 3 ( 𝑥3 3 + 𝑐) + ∫ −8𝑥3 𝑑𝑥 = 2𝑥 + 𝑐 + 3 ( 𝑥3 3 + 𝑐) ∫ −8∫ 𝑥3 𝑑𝑥 = 2𝑥 + 𝑐 + 3 ( 𝑥3 3 + 𝑐) − 8 ( 1 4 𝑥4 + 𝑐) = 2𝑥 + 𝑐 + 3 ( 𝑥3 3 + 𝑐) − 8 ( 𝑥4 4 + 𝑐) = 2𝑥 + 𝑥3 − 8 𝑥4 4 + 𝑐 = 2𝑥 + 𝑥3 + −8 1 𝑥4 4 + 𝑐 = 2𝑥 + 𝑥3 + −8𝑥4 4 + 𝑐
4.
= 2𝑥 +
𝑥3 − 2𝑥4 + 𝑐
5.
∫ (4𝑥3 -3𝑥2 + 6𝑥
− 1)dx = ∫ 4𝑥3 𝑑𝑥 + ∫ −3𝑥2 𝑑𝑥 + ∫ −1𝑑𝑥 = 4 ∫ 4𝑥3 𝑑𝑥 + ∫ −3𝑥2 𝑑𝑥 + ∫ 6𝑥𝑑𝑥 + ∫ −1𝑑𝑥 = 4 ( 1 4 𝑥4 1 + 𝑐) + ∫ −3𝑥2 𝑑𝑥 + ∫ 6𝑥𝑑𝑥 + ∫ −1𝑑𝑥 = 4 ( 𝑥4 4 + 𝑐) + ∫ −3𝑥2 𝑑𝑥 + ∫ 6𝑥𝑑𝑥 + ∫ −1𝑑𝑥 = 4 ( 𝑥4 4 + 𝑐) + −3 ∫ 𝑥2 𝑑𝑥 + ∫ 6𝑥𝑑𝑥 + ∫ −1𝑑𝑥 = 4 ( 𝑥4 4 + 𝑐) + −3 ∫ ( 1 3 𝑥3 + 𝑐) ∫ 6𝑥𝑑𝑥 + ∫ −1𝑑𝑥 = 4 ( 𝑥4 4 + 𝑐) + −3 ∫ ( 𝑥3 3 + 𝑐) ∫ 6𝑥𝑑𝑥 + ∫ −1𝑑𝑥 = 4 ( 𝑥4 4 + 𝑐) + −3 ∫ ( 𝑥3 3 + 𝑐) + 6 ∫ 𝑥𝑑𝑥 + ∫ −1𝑑𝑥 = 4 ( 𝑥4 4 + 𝑐) + −3 ∫ ( 𝑥3 3 + 𝑐) + 6 ∫( 1 2 𝑥2 + 𝑐) ∫ −1𝑑𝑥 = 4 ( 𝑥4 4 + 𝑐) − 3 ( 𝑥3 3 + 𝑐) + 6 ( 𝑥2 2 + 𝑐) ∫ −1𝑑𝑥 = 4 ( 𝑥4 4 + 𝑐) − 3 ( 𝑥3 3 + 𝑐) + 6 ( 𝑥2 2 + 𝑐) + (−𝑥 + 2) =𝑥4 − 𝑥3 + 3𝑥2 − 𝑥 + 𝑐 ∫(2𝑥2 − 4𝑥 + 4) = ∫ 2𝑥2 𝑑𝑥 + ∫ −4𝑥𝑑𝑥 + ∫ 4𝑑𝑥 = 2 ∫ 𝑥2 𝑑𝑥 + ∫ −4𝑥𝑑𝑥 + ∫ 4𝑑𝑥 = 2 ∫ 1 3 𝑥3 + 𝑐 ∫ −4𝑥𝑑𝑥 + ∫ 4𝑑𝑥 = 2 ( 𝑥3 3 + 𝑐) + ∫ −4𝑑𝑥 + ∫ 4𝑑𝑥 = 2 ( 𝑥3 3 + 𝑐) − 4 ( 1 2 𝑥2 + 𝑐) + ∫ 4𝑑𝑥 = 2 ( 𝑥3 3 + 𝑐) − 4 ( 𝑥2 2 + 𝑐) + ∫ 4𝑑𝑥 = 2 ( 𝑥3 3 + 𝑐) − 4 ( 𝑥2 2 + 𝑐) + (4𝑥 + 𝑐) = 2𝑥3 3 − 2𝑥2 + 4𝑥 + 𝑐 ∫ (3𝑥5 − 2𝑥3 ) dx = ∫ 3𝑥5 𝑑𝑥 + ∫ −2𝑥3 𝑑𝑥
6.
= 3 ∫
𝑥5 𝑑𝑥 + ∫ −2𝑥3 𝑑𝑥 = 3 ( 1 6 𝑥6 + 𝑐) + ∫ −2𝑥3 𝑑𝑥 = 3 ( 1 6 𝑥6 + 𝑐) − 2 ∫ 𝑥3 𝑑𝑥 = 3 ( 1 6 𝑥6 + 𝑐) − 2 ( 1 4 𝑥4 + 𝑐) = 𝑥6 2 + 2 𝑥4 4 + 𝑐 = 𝑥6 2 + 2(−𝑥4) 2(2) + 𝑐 = 𝑥6 2 + −𝑥4) 2 + 𝑐 = 𝑥6 2 − −𝑥4) 2 + 𝑐 = 1 2 𝑥6 + ( 1 2 𝑥4 + 𝑐) = 1 2 𝑥6 − 1 2 𝑥4 + 𝑐 5-∫ (3 − 2𝑥 + 𝑥2 ) dx = ∫ 3𝑑𝑥 + ∫ −2𝑥𝑑𝑥 + ∫ 𝑥2 𝑑𝑥 = 3𝑥 + 𝑐 ∫ −2𝑥𝑑𝑥 + ∫ 𝑥2 𝑑𝑥 = 3𝑥 + 𝑐 − 2 ∫ 𝑥𝑑𝑥 + ∫ 𝑥2 𝑑𝑥 = 3𝑥 + 𝑐 − 2 ( 1 2 𝑥2 + 𝑐) + ∫ 𝑥2 𝑑𝑥 = 3𝑥 + 𝑐 − 2 ( 𝑥2 2 + 𝑐) + ∫ 𝑥2 𝑑𝑥 = 3𝑥 + 𝑐 − 2 ( 𝑥2 2 + 𝑐) + ( 1 3 𝑥3 + 𝑐) = 3𝑥 + 𝑐 − 2 ( 𝑥2 2 + 𝑐) + 𝑥3 3 + 𝑐 = 3𝑥 − 𝑥2 + 𝑥3 3 + 𝑐 = 3𝑥 − 𝑥2 + 1 3 𝑥3 + 𝑐 FORMULA 4
7.
∫ 18𝑋9 𝐷𝑋 = =
18 ( 𝑋9+1 9 + 1 ) 𝐷𝑋 = 18 10 𝑋10 𝐷𝑋 ∫ 12𝑋6 𝐷𝑋 = = 12 ( 𝑋6+1 6 + 1 ) 𝐷𝑋 = 12 7 𝑋10 𝐷𝑋 ∫ 9𝑋8 𝐷𝑋 = = 9 ( 𝑋8+1 8 + 1 ) 𝐷𝑋 =1𝑋9 𝐷𝑋 ∫ 24𝑋7 𝐷𝑋 = = 24 ( 𝑋7+1 7 + 1 ) 𝐷𝑋 =3𝑋8 𝐷𝑋 ∫ 18𝑋2 𝐷𝑋 = = 18 ( 𝑋2+1 2 + 1 ) 𝐷𝑋 =6𝑋3 𝐷𝑋 FORMULA 5 ∫ 𝑉 𝑁 𝐷𝑉 = 𝑉 𝑁+1 𝑁 + 1
8.
∫ (𝑋5 − 2𝑋)2(𝑋 −
1)𝐷𝑋 = = 1 5 (𝑋5 − 4𝑋)5(𝑋 − 1)𝐷𝑋 = 1 5 (𝑋5 − 2𝑋)3 3 + 𝐶 = (𝑋5−5𝑋) 3 15 + 𝐶 ∫ (𝑋4 − 3𝑋)4(𝑋 − 2)𝐷𝑋 = = 1 4 (𝑋4 − 3𝑋)4(𝑋 − 2)𝐷𝑋 = 1 4 (𝑋7 − 3𝑋)5 5 + 𝐶 = (𝑋4−3𝑋)9 20 + 𝐶 ∫ (𝑋6 − 2𝑋)9(𝑋 − 1)𝐷𝑋 = = 1 6 (𝑋6 − 2𝑋)9(𝑋 − 1)𝐷𝑋 = 1 6 (𝑋6 − 2𝑋)10 4 + 𝐶 = (𝑋6−2𝑋)10 60 + 𝐶 ∫ (𝑋8 − 6𝑋)7(𝑋 − 1)𝐷𝑋 = = 1 8 (𝑋8 − 6𝑋)7(𝑋 − 1)𝐷𝑋 = 1 8 (𝑋8 − 6𝑋)8 8 + 𝐶
9.
= (𝑋8−6𝑋)8 64 + 𝐶 ∫ (𝑋3 − 16𝑋)5(𝑋
− 1)𝐷𝑋 = = 1 3 (𝑋3 − 16𝑋)5(𝑋 − 1)𝐷𝑋 = 1 3 (𝑋3 − 16𝑋)6 6 + 𝐶 = (𝑋3−16𝑋)6 18 + 𝐶
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