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Question text
Queremos integrar la función <span class="nolink">«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»f«/mi»«mo»(«/mo»«mi»x«/mi»«mo»)«/mo»«mo»=«/mo»«mi»#p«/mi»«mo»·«/mo»«mi»#q«/mi»«/math»</span>. Escoge todas las opciones correctas:<br />
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General feedback
<span style="font-weight: bold;">Comentario:</span><br /><br />Una vez termines el proceso de integración por partes debes obtener #r+C.<br /><br /><br />
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Algorithm
Choice 1
Answer
Podemos integrar por partes haciendo <math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>u</mi><mo>=</mo><mi>#q</mi></mrow></math> i <math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>dv</mi><mo>=</mo><mi>#p</mi></mrow></math>.
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Feedback
Efectivamente, esta es la forma correcta de escoger las funciones para integrar por partes.
Choice 2
Answer
Podemos integrar por partes haciendo <math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>u</mi><mo>=</mo><mi>#p</mi></mrow></math> i <math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>dv</mi><mo>=</mo><mi>#q</mi></mrow></math>.
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Cuidad que debes escoger las funciones al revés u=#q i dv=#p, porque tal como lo haces no podras integrar.
Choice 3
Answer
La derivada de #q es proporcional a la derivada de la función ln(x).
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Efectivamente la derivada de #q es #dq, y es igual a un número por la derivada de ln(x), que es 1/x.
Choice 4
Answer
Si integramos por partes, al final del proceso tan solo tendremos que integrar un monomio de grado #a.
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Correcto.
Choice 5
Answer
Si integramos por partes, al final del proceso tan solo tendremos que integrar un monomio de grado #b.
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Feedback
Vigila, porque cuando integras por partes, la derivada de u=#q es du=#dq. Si lo multiplicas por v=#v (ya que dv=#p) te queda #s, que es lo que debes integrar para acabar.
Choice 6
Answer
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Choice 7
Answer
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Choice 8
Answer
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Wednesday, 12 June 2013, 1:11 PM
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Wednesday, 12 June 2013, 1:11 PM
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