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Calculo integral daniel rangel

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Daniel Rangel
Daniel Rangel

con la ayuda de estos ejemplos podr叩n realizar anti-derivadas con las formulas ya dadas en el documento

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DANIEL RANGEL MEDINA 30 DE SEPTIEMBRE DE 2018 EDGAR MATA 4尊B
1. Introducci坦n al c叩lculointegral ylasanti derivadas La antiderivadaeslaoperaci坦ninversaa la derivada,坦seala antiderivadavaa deshacerloque la derivadase encargade hacer. En este documento elaboraremos algunos problemas bajo el formulario compartido en clase por el profesorEdgarMata. Aplicaci坦nde laf坦rmula1 Formula: = + Problema1: ( 6) = 2 2 2 + = 6 = 2 Problema2: (83 92 + 4) 8 3 モ 9モ ヰ+ 4 84 4 93 3 + 4 + Problemas3:(23 12 3) 2 3 モ 12 ヰ 3 24 4 122 2 3 + Problema4: 43 = 4 4 44 4 + Problema5: (163 182 + 8) 16 3 モ 18 ヰ+ 8 164 4 183 3 + 8 + 2. Aplicaci坦nde laf坦rmula2 Formula: = +1 +1 + Problema1: (46 82 + 6) 4 6 8 2 + 6 ＄
47 7 83 3 + 62 2 + Problema2:(128 104 6) 12 8 10 4 6 ＄ 129 9 105 5 62 2 + Problema3: (82) = 8 2 = 83 3 + Problema4: (205 156 + 5) 20 5 15 6+5 + 5 126 6 157 7 + 5 2 + Problema5:(205 156 + 5) 20 5 15 6 + 5 126 6 157 7 + 52 2 + 3. Aplicaci坦nde laf坦rmula3: ( + ) = + Problema1: (52 + 52 3) 52 + 5 5 3 53 3 + 52 2 3 + Problema2:(122 + 244 6) 122 + 24 24 6 123 3 + 245 5 6 + Problema3:(1610 + 168 18) 1610 + 16 16 18 1611 11 + 169 9 18 +
Problema4:2015 + 1510 10 2015 + 15 15 10 2016 16 + 1511 11 10 + Problema5:10020 + 55100 250 10021 21 55101 101 250 + 4. Aplicaci坦nde laf坦rmula4 Formula: = Problema1: ( 3 2 + 2 + 1) 1 3 2 + 2 + 1 4 2 4 2 + 22 2 + + Problema2: (83 92 + 4) 8 3 9 2 + 4 ヰ 84 4 93 3 + 4 Problema3: ( 4 + 22 + 1) 1 4 + 2 2 + 1 ＄ 5 5 + 23 3 + Problema4: ( + 2) 1 + 2 ヰ 2 2 + 2 + Problema5: (34 52 + ) 3 4 5 2 + 1 ヰ 35 5 53 3 + +
5. Aplicaci坦nde laf坦rmula5 F坦rmula5: 5 = +1 +1 + Problema 1: ( + 1) (34 2)2 = (34 2)2( + 1) Problema2: ( + 2) (66 6)6 = (66 6)6( + 2) Problema3: ( + 12) (246 + 3)32 = (246 + 3)32( + 12) V= 34 2 = (123 2) = 1 2 (34 2)2 2(63 ) = 1 2 (34 2)3 3 + = 1 6 (34 2)3 + V= 66 6 = (366 6) = 6(65 1) = 1 6 (66 6)7 6(65 1) = 1 6 (66 2)7 3 + = 1 42 (66 6)7 + = 1 3 (246 + 3)33 3(485 1) = 1 3 (246 + 3)33 33 + = 1 99 (246 3)33 + V= 246 + 3 = (1445 3) = 3(485 1)
Problema4: ( + 4) (32 + 42 )6 = (32 + 42 )6( + 4) Problema5: ( + 6) (98 + 1310 )8 = (98 + 1610 )32( + 6) 6. Aplicaci坦nde laf坦rmula6 Formula: = + 1= + = Problema1: ヰ (3+3) = 2ヰ 2+5 Problema2: ( +2) (3+3) V= 32 + 42 = (6 + 8) = 2(3 + 4) = 1 2 (32 + 42)7 2(3 4) = 1 14 (32 + 42)7 7 + = 1 14 (32 42)7 + V= 98 + 1610 = (727 + 1309) = 8(97 + 16) = 1 8 (98 + 1610)33 8(97 + 16) = 1 8 (98 + 1610)33 33 + = 1 264 (98 + 1610)33 + V= 98 + 1610 = (2) = 1 2 | 2 + 5| + V= 3 + 3 = (3 + 3) = 3( + 1) ヰ 1 3 3( + 1) = 1 3 | 3 + 3| +
Problema3: 4 9 = 4 モ9 = 4モ8 8 + = 2モ8 + = 2 8 + Problema4: 4 23+6 = 1 2 62 23+6 = 1 2 |23 + 6| + Problema5: 1 5 = 1 5 Encabezados: Los problemasfueronextra鱈dosde lasiguientebibliograf鱈a = 23 + 6 = 62 = 1モ4 4 + = 4 モ4 +
Calculo integral daniel rangel

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Calculo integral daniel rangel

  • 1. DANIEL RANGEL MEDINA 30 DE SEPTIEMBRE DE 2018 EDGAR MATA 4尊B
  • 2. 1. Introducci坦n al c叩lculointegral ylasanti derivadas La antiderivadaeslaoperaci坦ninversaa la derivada,坦seala antiderivadavaa deshacerloque la derivadase encargade hacer. En este documento elaboraremos algunos problemas bajo el formulario compartido en clase por el profesorEdgarMata. Aplicaci坦nde laf坦rmula1 Formula: = + Problema1: ( 6) = 2 2 2 + = 6 = 2 Problema2: (83 92 + 4) 8 3 モ 9モ ヰ+ 4 84 4 93 3 + 4 + Problemas3:(23 12 3) 2 3 モ 12 ヰ 3 24 4 122 2 3 + Problema4: 43 = 4 4 44 4 + Problema5: (163 182 + 8) 16 3 モ 18 ヰ+ 8 164 4 183 3 + 8 + 2. Aplicaci坦nde laf坦rmula2 Formula: = +1 +1 + Problema1: (46 82 + 6) 4 6 8 2 + 6 ＄
  • 3. 47 7 83 3 + 62 2 + Problema2:(128 104 6) 12 8 10 4 6 ＄ 129 9 105 5 62 2 + Problema3: (82) = 8 2 = 83 3 + Problema4: (205 156 + 5) 20 5 15 6+5 + 5 126 6 157 7 + 5 2 + Problema5:(205 156 + 5) 20 5 15 6 + 5 126 6 157 7 + 52 2 + 3. Aplicaci坦nde laf坦rmula3: ( + ) = + Problema1: (52 + 52 3) 52 + 5 5 3 53 3 + 52 2 3 + Problema2:(122 + 244 6) 122 + 24 24 6 123 3 + 245 5 6 + Problema3:(1610 + 168 18) 1610 + 16 16 18 1611 11 + 169 9 18 +
  • 4. Problema4:2015 + 1510 10 2015 + 15 15 10 2016 16 + 1511 11 10 + Problema5:10020 + 55100 250 10021 21 55101 101 250 + 4. Aplicaci坦nde laf坦rmula4 Formula: = Problema1: ( 3 2 + 2 + 1) 1 3 2 + 2 + 1 4 2 4 2 + 22 2 + + Problema2: (83 92 + 4) 8 3 9 2 + 4 ヰ 84 4 93 3 + 4 Problema3: ( 4 + 22 + 1) 1 4 + 2 2 + 1 ＄ 5 5 + 23 3 + Problema4: ( + 2) 1 + 2 ヰ 2 2 + 2 + Problema5: (34 52 + ) 3 4 5 2 + 1 ヰ 35 5 53 3 + +
  • 5. 5. Aplicaci坦nde laf坦rmula5 F坦rmula5: 5 = +1 +1 + Problema 1: ( + 1) (34 2)2 = (34 2)2( + 1) Problema2: ( + 2) (66 6)6 = (66 6)6( + 2) Problema3: ( + 12) (246 + 3)32 = (246 + 3)32( + 12) V= 34 2 = (123 2) = 1 2 (34 2)2 2(63 ) = 1 2 (34 2)3 3 + = 1 6 (34 2)3 + V= 66 6 = (366 6) = 6(65 1) = 1 6 (66 6)7 6(65 1) = 1 6 (66 2)7 3 + = 1 42 (66 6)7 + = 1 3 (246 + 3)33 3(485 1) = 1 3 (246 + 3)33 33 + = 1 99 (246 3)33 + V= 246 + 3 = (1445 3) = 3(485 1)
  • 6. Problema4: ( + 4) (32 + 42 )6 = (32 + 42 )6( + 4) Problema5: ( + 6) (98 + 1310 )8 = (98 + 1610 )32( + 6) 6. Aplicaci坦nde laf坦rmula6 Formula: = + 1= + = Problema1: ヰ (3+3) = 2ヰ 2+5 Problema2: ( +2) (3+3) V= 32 + 42 = (6 + 8) = 2(3 + 4) = 1 2 (32 + 42)7 2(3 4) = 1 14 (32 + 42)7 7 + = 1 14 (32 42)7 + V= 98 + 1610 = (727 + 1309) = 8(97 + 16) = 1 8 (98 + 1610)33 8(97 + 16) = 1 8 (98 + 1610)33 33 + = 1 264 (98 + 1610)33 + V= 98 + 1610 = (2) = 1 2 | 2 + 5| + V= 3 + 3 = (3 + 3) = 3( + 1) ヰ 1 3 3( + 1) = 1 3 | 3 + 3| +
  • 7. Problema3: 4 9 = 4 モ9 = 4モ8 8 + = 2モ8 + = 2 8 + Problema4: 4 23+6 = 1 2 62 23+6 = 1 2 |23 + 6| + Problema5: 1 5 = 1 5 Encabezados: Los problemasfueronextra鱈dosde lasiguientebibliograf鱈a = 23 + 6 = 62 = 1モ4 4 + = 4 モ4 +