Ayuda por favor con estos ejercicios de calculo integral. Con procedimiento y propiedad aplicada en los ejercicios. gracias.
Halle el área bajo la curva de las siguientes funciones: (Visualizar y Ampliar Imagen)
Graficar cada uno de los puntos de las funciones.
Adjuntos:
![](https://es-static.z-dn.net/files/d2b/3cc23398f5c27b2f7969dcee7c42c6f2.jpg)
Respuestas
Respuesta dada por:
3
De igual manera por motivo de que me hes permitido subir un máximo de 5 imágenes no subiré la imágenes correspondientes aunque, no representan ayuda alguna ...el propósito de los siguientes ejercicio es hallar el área bajo la curva usando los límites establecidos...y usando el Teorema Fundamental del Cálculo
![\int\limits^b_a {f(x)} \, dx =F(b)-F(a) \int\limits^b_a {f(x)} \, dx =F(b)-F(a)](https://tex.z-dn.net/?f=+%5Cint%5Climits%5Eb_a+%7Bf%28x%29%7D+%5C%2C+dx+%3DF%28b%29-F%28a%29)
Siendo F mayúscula la primitiva de cada integral...Bueno empecemos:
(1)
![\int\limits^1_0{(2x-3)} \, dx = 2\int\limits^1_0 {x} \, dx -3 \int\limits^1_0 {} \, dx = x^{2} -3x]\limits^1_0 \\ \\ F(b)-F(a) \\ ( (1)^{2} -3(1))-( (0)^{2}-3(0) ) \\ -2 \int\limits^1_0{(2x-3)} \, dx = 2\int\limits^1_0 {x} \, dx -3 \int\limits^1_0 {} \, dx = x^{2} -3x]\limits^1_0 \\ \\ F(b)-F(a) \\ ( (1)^{2} -3(1))-( (0)^{2}-3(0) ) \\ -2](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E1_0%7B%282x-3%29%7D+%5C%2C+dx+%3D++2%5Cint%5Climits%5E1_0+%7Bx%7D+%5C%2C+dx+-3+%5Cint%5Climits%5E1_0+%7B%7D+%5C%2C+dx+%3D+x%5E%7B2%7D+-3x%5D%5Climits%5E1_0++%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C+%28+%281%29%5E%7B2%7D+-3%281%29%29-%28+%280%29%5E%7B2%7D-3%280%29+%29+%5C%5C+-2)
(2)
![\int\limits^2_1 { \frac{5-x}{ x^{3} } } \, dx = \int\limits^2_1 { \frac{5}{ x^{3} } } \, dx- \int\limits^2_1 { \frac{1}{ x^{2} } } \, dx = (- \frac{5}{2 x^{2} } + \frac{1}{x})]\limits^2_1 \\ \\ F(b)-F(a) \\ (- \frac{5}{2 (2)^{2} } + \frac{1}{(2)})-((- \frac{5}{2 (1)^{2} } + \frac{1}{(1)})) \\ -\frac{1}{8} + \frac{3}{2} = \frac{11}{8} =1,375 \int\limits^2_1 { \frac{5-x}{ x^{3} } } \, dx = \int\limits^2_1 { \frac{5}{ x^{3} } } \, dx- \int\limits^2_1 { \frac{1}{ x^{2} } } \, dx = (- \frac{5}{2 x^{2} } + \frac{1}{x})]\limits^2_1 \\ \\ F(b)-F(a) \\ (- \frac{5}{2 (2)^{2} } + \frac{1}{(2)})-((- \frac{5}{2 (1)^{2} } + \frac{1}{(1)})) \\ -\frac{1}{8} + \frac{3}{2} = \frac{11}{8} =1,375](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E2_1+%7B+%5Cfrac%7B5-x%7D%7B+x%5E%7B3%7D+%7D+%7D+%5C%2C+dx+%3D+%5Cint%5Climits%5E2_1+%7B+%5Cfrac%7B5%7D%7B+x%5E%7B3%7D+%7D+%7D+%5C%2C+dx-+%5Cint%5Climits%5E2_1+%7B+%5Cfrac%7B1%7D%7B+x%5E%7B2%7D+%7D+%7D+%5C%2C+dx+%3D+%28-+%5Cfrac%7B5%7D%7B2+x%5E%7B2%7D+%7D+%2B+%5Cfrac%7B1%7D%7Bx%7D%29%5D%5Climits%5E2_1+%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C+%28-+%5Cfrac%7B5%7D%7B2+%282%29%5E%7B2%7D+%7D+%2B+%5Cfrac%7B1%7D%7B%282%29%7D%29-%28%28-+%5Cfrac%7B5%7D%7B2+%281%29%5E%7B2%7D+%7D+%2B+%5Cfrac%7B1%7D%7B%281%29%7D%29%29+%5C%5C++-%5Cfrac%7B1%7D%7B8%7D+%2B+%5Cfrac%7B3%7D%7B2%7D+%3D+%5Cfrac%7B11%7D%7B8%7D+%3D1%2C375)
(3)
![\int\limits^5_1 {2 \sqrt{x-1} } \, dx =2 \int\limits^5_1 {(x-1)^{ \frac{1}{2} } } \, dx =4( \frac{ (x-1)^{ \frac{3}{2} } }{3} )] \limits^5_1 \\ \\ F(b)-F(a) \\ 4( \frac{ ((5)-1)^{ \frac{3}{2} } }{3} )-(4( \frac{ ((1)-1)^{ \frac{3}{2} } }{3} )) \\ \frac{4}{3} ( 4^{ \frac{3}{2} } )=10,6667 \int\limits^5_1 {2 \sqrt{x-1} } \, dx =2 \int\limits^5_1 {(x-1)^{ \frac{1}{2} } } \, dx =4( \frac{ (x-1)^{ \frac{3}{2} } }{3} )] \limits^5_1 \\ \\ F(b)-F(a) \\ 4( \frac{ ((5)-1)^{ \frac{3}{2} } }{3} )-(4( \frac{ ((1)-1)^{ \frac{3}{2} } }{3} )) \\ \frac{4}{3} ( 4^{ \frac{3}{2} } )=10,6667](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E5_1+%7B2+%5Csqrt%7Bx-1%7D+%7D+%5C%2C+dx+%3D2+%5Cint%5Climits%5E5_1+%7B%28x-1%29%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7D+%7D+%5C%2C+dx+%3D4%28+%5Cfrac%7B+%28x-1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D+%7D%7B3%7D+%29%5D+%5Climits%5E5_1++%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C+4%28+%5Cfrac%7B+%28%285%29-1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D+%7D%7B3%7D+%29-%284%28+%5Cfrac%7B+%28%281%29-1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D+%7D%7B3%7D+%29%29+%5C%5C+++%5Cfrac%7B4%7D%7B3%7D+%28+4%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D+%29%3D10%2C6667)
(4)
![\int\limits^a_0 { ( \sqrt{a}- \sqrt{x} )^{2} } \, dx = \int\limits^a_0 {(a-2 \sqrt{a} \sqrt{x} +x )} \, dx = ax- \frac{4}{3} \sqrt{a} x^{ \frac{3}{2} } + \frac{ x^{2} }{2} ]\limits^a_0 \\ \\ F(b)-F(a) \\( a(a)- \frac{4}{3} \sqrt{a} (a)^{ \frac{3}{2} } + \frac{ (a)^{2} }{2})-(a(0)- \frac{4}{3} \sqrt{a} (0)^{ \frac{3}{2} } + \frac{ (0)^{2} }{2}) \\ a^{2} - \frac{4}{3} a^{2} + \frac{ a^{2} }{2} = \frac{ a^{2} }{6} \int\limits^a_0 { ( \sqrt{a}- \sqrt{x} )^{2} } \, dx = \int\limits^a_0 {(a-2 \sqrt{a} \sqrt{x} +x )} \, dx = ax- \frac{4}{3} \sqrt{a} x^{ \frac{3}{2} } + \frac{ x^{2} }{2} ]\limits^a_0 \\ \\ F(b)-F(a) \\( a(a)- \frac{4}{3} \sqrt{a} (a)^{ \frac{3}{2} } + \frac{ (a)^{2} }{2})-(a(0)- \frac{4}{3} \sqrt{a} (0)^{ \frac{3}{2} } + \frac{ (0)^{2} }{2}) \\ a^{2} - \frac{4}{3} a^{2} + \frac{ a^{2} }{2} = \frac{ a^{2} }{6}](https://tex.z-dn.net/?f=+%5Cint%5Climits%5Ea_0+%7B+%28+%5Csqrt%7Ba%7D-+%5Csqrt%7Bx%7D++%29%5E%7B2%7D+%7D+%5C%2C+dx+%3D+%5Cint%5Climits%5Ea_0+%7B%28a-2+%5Csqrt%7Ba%7D+%5Csqrt%7Bx%7D+%2Bx+%29%7D+%5C%2C+dx+%3D+ax-++%5Cfrac%7B4%7D%7B3%7D+%5Csqrt%7Ba%7D+x%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D++%2B+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D+%5D%5Climits%5Ea_0++%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C%28+a%28a%29-++%5Cfrac%7B4%7D%7B3%7D+%5Csqrt%7Ba%7D+%28a%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D++%2B+%5Cfrac%7B+%28a%29%5E%7B2%7D+%7D%7B2%7D%29-%28a%280%29-++%5Cfrac%7B4%7D%7B3%7D+%5Csqrt%7Ba%7D+%280%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D++%2B+%5Cfrac%7B+%280%29%5E%7B2%7D+%7D%7B2%7D%29+%5C%5C++a%5E%7B2%7D+-+%5Cfrac%7B4%7D%7B3%7D++a%5E%7B2%7D+%2B+%5Cfrac%7B+a%5E%7B2%7D+%7D%7B2%7D+%3D+%5Cfrac%7B+a%5E%7B2%7D+%7D%7B6%7D+)
(5)No me deja poner el signo "negativo" en los límites, aunque no éste escrito el límite inferior es "-2"
![\int\limits^0_2 {(x-2)(x+1)} \, dx = \int\limits^0_2 { x^{2} -x-2} \, dx = \frac{ x^{3} }{3} - \frac{ x^{2} }{2} -2x ] \limits^0_2 \\ \\ F(b)-F(a) \\ (\frac{ (0)^{3} }{3} - \frac{ (0)^{2} }{2} -2(0))-(\frac{ (-2)^{3} }{3} - \frac{ (-2)^{2} }{2} -2(-2) ) \\ -(- \frac{8}{3} -2+4)= \frac{2}{3} =0,6667 \int\limits^0_2 {(x-2)(x+1)} \, dx = \int\limits^0_2 { x^{2} -x-2} \, dx = \frac{ x^{3} }{3} - \frac{ x^{2} }{2} -2x ] \limits^0_2 \\ \\ F(b)-F(a) \\ (\frac{ (0)^{3} }{3} - \frac{ (0)^{2} }{2} -2(0))-(\frac{ (-2)^{3} }{3} - \frac{ (-2)^{2} }{2} -2(-2) ) \\ -(- \frac{8}{3} -2+4)= \frac{2}{3} =0,6667](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E0_2+%7B%28x-2%29%28x%2B1%29%7D+%5C%2C+dx+%3D+%5Cint%5Climits%5E0_2+%7B+x%5E%7B2%7D+-x-2%7D+%5C%2C+dx+%3D+%5Cfrac%7B+x%5E%7B3%7D+%7D%7B3%7D+-+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D+-2x+%5D+%5Climits%5E0_2+%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C+%28%5Cfrac%7B+%280%29%5E%7B3%7D+%7D%7B3%7D+-+%5Cfrac%7B+%280%29%5E%7B2%7D+%7D%7B2%7D+-2%280%29%29-%28%5Cfrac%7B+%28-2%29%5E%7B3%7D+%7D%7B3%7D+-+%5Cfrac%7B+%28-2%29%5E%7B2%7D+%7D%7B2%7D+-2%28-2%29+%29+%5C%5C+-%28-+%5Cfrac%7B8%7D%7B3%7D+-2%2B4%29%3D+%5Cfrac%7B2%7D%7B3%7D+%3D0%2C6667)
(6)
![\int\limits^4_0 { (1+2 \sqrt{x} )^{2} } \, dx = \int\limits^4_0 {1+4 \sqrt{x} +4x} \, dx =x+ \frac{8}{3} x^{ \frac{3}{2} } +2 x^{2}]\limits^4_0 \\ \\ F(b)-F(a) \\ \\ ((4)+ \frac{8}{3} (4)^{ \frac{3}{2} } +2 (4)^{2})-((0)+ \frac{8}{3} (0)^{ \frac{3}{2} } +2 (0)^{2}) \\ \\ 57,333 \int\limits^4_0 { (1+2 \sqrt{x} )^{2} } \, dx = \int\limits^4_0 {1+4 \sqrt{x} +4x} \, dx =x+ \frac{8}{3} x^{ \frac{3}{2} } +2 x^{2}]\limits^4_0 \\ \\ F(b)-F(a) \\ \\ ((4)+ \frac{8}{3} (4)^{ \frac{3}{2} } +2 (4)^{2})-((0)+ \frac{8}{3} (0)^{ \frac{3}{2} } +2 (0)^{2}) \\ \\ 57,333](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E4_0+%7B+%281%2B2+%5Csqrt%7Bx%7D+%29%5E%7B2%7D+%7D+%5C%2C+dx+%3D+%5Cint%5Climits%5E4_0+%7B1%2B4+%5Csqrt%7Bx%7D+%2B4x%7D+%5C%2C+dx+%3Dx%2B+%5Cfrac%7B8%7D%7B3%7D++x%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D+%2B2+x%5E%7B2%7D%5D%5Climits%5E4_0+%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C++%5C%5C+%28%284%29%2B+%5Cfrac%7B8%7D%7B3%7D++%284%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D+%2B2+%284%29%5E%7B2%7D%29-%28%280%29%2B+%5Cfrac%7B8%7D%7B3%7D++%280%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D+%2B2+%280%29%5E%7B2%7D%29+%5C%5C++%5C%5C+57%2C333)
(7)
![\int\limits^1_0 {(2a+1) ^{4} } \, da \\ Cosideremos: \\ u=2a+1 \\ du=2da \\ da= \frac{du}{2} \\ Reemplazando: \\ \int\limits^1_0 { u^{4} } \, \frac{du}{2} = \frac{(2a+1) ^{5} }{10}] \limits^1_0 \\ \\ F(b)-F(a) \\ \\ (\frac{(2(1)+1) ^{5} }{10})-(\frac{(2(0)+1) ^{5} }{10}) =24,2 \int\limits^1_0 {(2a+1) ^{4} } \, da \\ Cosideremos: \\ u=2a+1 \\ du=2da \\ da= \frac{du}{2} \\ Reemplazando: \\ \int\limits^1_0 { u^{4} } \, \frac{du}{2} = \frac{(2a+1) ^{5} }{10}] \limits^1_0 \\ \\ F(b)-F(a) \\ \\ (\frac{(2(1)+1) ^{5} }{10})-(\frac{(2(0)+1) ^{5} }{10}) =24,2](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E1_0+%7B%282a%2B1%29+%5E%7B4%7D+%7D+%5C%2C+da+%5C%5C+Cosideremos%3A+%5C%5C+u%3D2a%2B1+%5C%5C+du%3D2da+%5C%5C+da%3D+%5Cfrac%7Bdu%7D%7B2%7D+++%5C%5C+Reemplazando%3A+%5C%5C++%5Cint%5Climits%5E1_0+%7B+u%5E%7B4%7D+%7D+%5C%2C++%5Cfrac%7Bdu%7D%7B2%7D+%3D+%5Cfrac%7B%282a%2B1%29+%5E%7B5%7D+%7D%7B10%7D%5D+%5Climits%5E1_0++%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C++%5C%5C+%28%5Cfrac%7B%282%281%29%2B1%29+%5E%7B5%7D+%7D%7B10%7D%29-%28%5Cfrac%7B%282%280%29%2B1%29+%5E%7B5%7D+%7D%7B10%7D%29+%3D24%2C2)
(8)
![\int\limits^8_1 {xln(x)} \, dx = \frac{1}{2} x^{2} ln(x) - \frac{ x^{4} }{4}] \limits^8_1 \\ \\ F(b)-F(a) \\ \\ (\frac{1}{2} (8)^{2} ln((8)) - \frac{ (8)^{4} }{4})-(\frac{1}{2} (1)^{2} ln((1)) - \frac{ (1)^{4} }{4}) \\ 32ln(8)- \frac{63}{4} =50,79 \int\limits^8_1 {xln(x)} \, dx = \frac{1}{2} x^{2} ln(x) - \frac{ x^{4} }{4}] \limits^8_1 \\ \\ F(b)-F(a) \\ \\ (\frac{1}{2} (8)^{2} ln((8)) - \frac{ (8)^{4} }{4})-(\frac{1}{2} (1)^{2} ln((1)) - \frac{ (1)^{4} }{4}) \\ 32ln(8)- \frac{63}{4} =50,79](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E8_1+%7Bxln%28x%29%7D+%5C%2C+dx+%3D+%5Cfrac%7B1%7D%7B2%7D+x%5E%7B2%7D+ln%28x%29+-+%5Cfrac%7B+x%5E%7B4%7D+%7D%7B4%7D%5D+%5Climits%5E8_1+%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C++%5C%5C+%28%5Cfrac%7B1%7D%7B2%7D+%288%29%5E%7B2%7D+ln%28%288%29%29+-+%5Cfrac%7B+%288%29%5E%7B4%7D+%7D%7B4%7D%29-%28%5Cfrac%7B1%7D%7B2%7D+%281%29%5E%7B2%7D+ln%28%281%29%29+-+%5Cfrac%7B+%281%29%5E%7B4%7D+%7D%7B4%7D%29+%5C%5C+32ln%288%29-+%5Cfrac%7B63%7D%7B4%7D+%3D50%2C79)
(9)
![\int\limits^ \frac{ \pi }{2} _0 {sin(x)} \, dx =-cos(x)]\limits^ \frac{ \pi }{2} _0 \\ \\ F(b) -F(a) \\ \\ (-cos( \frac{ \pi }{2} ))-(-cos(0)) \\ -(-1)=1 \int\limits^ \frac{ \pi }{2} _0 {sin(x)} \, dx =-cos(x)]\limits^ \frac{ \pi }{2} _0 \\ \\ F(b) -F(a) \\ \\ (-cos( \frac{ \pi }{2} ))-(-cos(0)) \\ -(-1)=1](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+_0+%7Bsin%28x%29%7D+%5C%2C+dx+%3D-cos%28x%29%5D%5Climits%5E+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+_0+%5C%5C++%5C%5C+F%28b%29+-F%28a%29+%5C%5C++%5C%5C+%28-cos%28+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%29%29-%28-cos%280%29%29+%5C%5C+-%28-1%29%3D1)
(10)
![\int\limits^3_0 {- x^{2} +x-1} \, dx =- \frac{ x^{3} }{3} + \frac{ x^{2} }{2} -x ] \limits^3_0 \\ \\ F(b)-F(a) \\ \\ (- \frac{ (3)^{3} }{3} + \frac{ (3)^{2} }{2} -(3))-(- \frac{ (0)^{3} }{3} + \frac{ (0)^{2} }{2} -(0)) \\ -12+ \frac{9}{2} =- \frac{15}{2} \int\limits^3_0 {- x^{2} +x-1} \, dx =- \frac{ x^{3} }{3} + \frac{ x^{2} }{2} -x ] \limits^3_0 \\ \\ F(b)-F(a) \\ \\ (- \frac{ (3)^{3} }{3} + \frac{ (3)^{2} }{2} -(3))-(- \frac{ (0)^{3} }{3} + \frac{ (0)^{2} }{2} -(0)) \\ -12+ \frac{9}{2} =- \frac{15}{2}](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E3_0+%7B-+x%5E%7B2%7D+%2Bx-1%7D+%5C%2C+dx+%3D-+%5Cfrac%7B+x%5E%7B3%7D+%7D%7B3%7D+%2B+%5Cfrac%7B+x%5E%7B2%7D+%7D%7B2%7D+-x+%5D+%5Climits%5E3_0+%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C++%5C%5C+%28-+%5Cfrac%7B+%283%29%5E%7B3%7D+%7D%7B3%7D+%2B+%5Cfrac%7B+%283%29%5E%7B2%7D+%7D%7B2%7D+-%283%29%29-%28-+%5Cfrac%7B+%280%29%5E%7B3%7D+%7D%7B3%7D+%2B+%5Cfrac%7B+%280%29%5E%7B2%7D+%7D%7B2%7D+-%280%29%29+%5C%5C+-12%2B+%5Cfrac%7B9%7D%7B2%7D+%3D-+%5Cfrac%7B15%7D%7B2%7D+)
(11)Igualmente no puedo poner el signo negativo..pero tu sabes que está ahí
![\int\limits^1_2 {( \frac{1}{3}t-2 ) ^{2} } \, dt= \int\limits^1_2 { \frac{ t^{2} }{9}- \frac{4}{3}t +4 } \, dt= \frac{ t^{3} }{27} - \frac{2}{3} t^{2} +4t ]\limits^1_2 \\ \\ F(b)-F(a) \\ \\ (\frac{ (1)^{3} }{27} - \frac{2}{3} (1)^{2} +4(1) )-(\frac{ (-2)^{3} }{27} - \frac{2}{3} (-2)^{2} +4(-2) ) = \frac{43}{3} =14,333 \int\limits^1_2 {( \frac{1}{3}t-2 ) ^{2} } \, dt= \int\limits^1_2 { \frac{ t^{2} }{9}- \frac{4}{3}t +4 } \, dt= \frac{ t^{3} }{27} - \frac{2}{3} t^{2} +4t ]\limits^1_2 \\ \\ F(b)-F(a) \\ \\ (\frac{ (1)^{3} }{27} - \frac{2}{3} (1)^{2} +4(1) )-(\frac{ (-2)^{3} }{27} - \frac{2}{3} (-2)^{2} +4(-2) ) = \frac{43}{3} =14,333](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E1_2+%7B%28+%5Cfrac%7B1%7D%7B3%7Dt-2+%29+%5E%7B2%7D+%7D+%5C%2C+dt%3D+%5Cint%5Climits%5E1_2+%7B+%5Cfrac%7B+t%5E%7B2%7D+%7D%7B9%7D-+%5Cfrac%7B4%7D%7B3%7Dt+%2B4+%7D+%5C%2C+dt%3D+%5Cfrac%7B+t%5E%7B3%7D+%7D%7B27%7D+-+%5Cfrac%7B2%7D%7B3%7D+t%5E%7B2%7D++%2B4t+%5D%5Climits%5E1_2++%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C++%5C%5C+%28%5Cfrac%7B+%281%29%5E%7B3%7D+%7D%7B27%7D+-+%5Cfrac%7B2%7D%7B3%7D+%281%29%5E%7B2%7D++%2B4%281%29+%29-%28%5Cfrac%7B+%28-2%29%5E%7B3%7D+%7D%7B27%7D+-+%5Cfrac%7B2%7D%7B3%7D+%28-2%29%5E%7B2%7D++%2B4%28-2%29+%29+%3D+%5Cfrac%7B43%7D%7B3%7D+%3D14%2C333)
(12)de la misma manera no me deja poner el signo negativo pero sabes que está ahí en el límite inferior
![\int\limits^0_3 { \frac{1}{9+2x} } \, dx \\ Consideremos: \\ u=9+2x \\ du=2dx \\ dx= \frac{du}{2} \\ Reemplazamos: \\ \int\limits^0_3 { \frac{1}{u} } \, \frac{du}{2} = \frac{1}{2} ln(9+2x) \\ \\ F(b)-F(a) \\ \\ (\frac{1}{2} ln(9+2(0)))-(\frac{1}{2} ln(9+2(-3)))=0,255 \int\limits^0_3 { \frac{1}{9+2x} } \, dx \\ Consideremos: \\ u=9+2x \\ du=2dx \\ dx= \frac{du}{2} \\ Reemplazamos: \\ \int\limits^0_3 { \frac{1}{u} } \, \frac{du}{2} = \frac{1}{2} ln(9+2x) \\ \\ F(b)-F(a) \\ \\ (\frac{1}{2} ln(9+2(0)))-(\frac{1}{2} ln(9+2(-3)))=0,255](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E0_3+%7B+%5Cfrac%7B1%7D%7B9%2B2x%7D+%7D+%5C%2C+dx++%5C%5C+Consideremos%3A+%5C%5C+u%3D9%2B2x+%5C%5C+du%3D2dx+%5C%5C+dx%3D+%5Cfrac%7Bdu%7D%7B2%7D++%5C%5C+Reemplazamos%3A+%5C%5C++%5Cint%5Climits%5E0_3+%7B+%5Cfrac%7B1%7D%7Bu%7D+%7D+%5C%2C++%5Cfrac%7Bdu%7D%7B2%7D+%3D+%5Cfrac%7B1%7D%7B2%7D+ln%289%2B2x%29+%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C++%5C%5C+%28%5Cfrac%7B1%7D%7B2%7D+ln%289%2B2%280%29%29%29-%28%5Cfrac%7B1%7D%7B2%7D+ln%289%2B2%28-3%29%29%29%3D0%2C255)
(13)
![\int\limits^ \pi _0 {cos(2x)} \, dx = \frac{sin(2x)}{2}] \limits^ \pi _0 \\ \\ F(b)-F(a) \\ \\ (\frac{sin(2( \pi ))}{2})-(\frac{sin(2(0))}{2}) \\ \\ 0-0=0 \int\limits^ \pi _0 {cos(2x)} \, dx = \frac{sin(2x)}{2}] \limits^ \pi _0 \\ \\ F(b)-F(a) \\ \\ (\frac{sin(2( \pi ))}{2})-(\frac{sin(2(0))}{2}) \\ \\ 0-0=0](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E+%5Cpi+_0+%7Bcos%282x%29%7D+%5C%2C+dx+%3D+%5Cfrac%7Bsin%282x%29%7D%7B2%7D%5D+%5Climits%5E+%5Cpi+_0+%5C%5C++%5C%5C+F%28b%29-F%28a%29+%5C%5C++%5C%5C+%28%5Cfrac%7Bsin%282%28+%5Cpi+%29%29%7D%7B2%7D%29-%28%5Cfrac%7Bsin%282%280%29%29%7D%7B2%7D%29+%5C%5C++%5C%5C+0-0%3D0)
En éste caso lo que sucede es que las áreas que consideramos son las mismas por eso se cancelan..las áreas son =3,134
Espero te sirva
Siendo F mayúscula la primitiva de cada integral...Bueno empecemos:
(1)
(2)
(3)
(4)
(5)No me deja poner el signo "negativo" en los límites, aunque no éste escrito el límite inferior es "-2"
(6)
(7)
(8)
(9)
(10)
(11)Igualmente no puedo poner el signo negativo..pero tu sabes que está ahí
(12)de la misma manera no me deja poner el signo negativo pero sabes que está ahí en el límite inferior
(13)
En éste caso lo que sucede es que las áreas que consideramos son las mismas por eso se cancelan..las áreas son =3,134
Espero te sirva
Mafisterv:
Muchas Gracias
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