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Para calcular el área cerrada por la gráfica de <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»#f«/mi»«/math»</span> y el eje de las abscisas en el intervalo <span class="nolink">«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfenced close=¨]¨ open=¨[¨»«mrow»«mi»#a«/mi»«mo»,«/mo»«mi»#c«/mi»«/mrow»«/mfenced»«/math»</span>se debe resolver:
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Choice 1
Answer
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msubsup»«mo»§#8747;«/mo»«mi»#a«/mi»«mi»#c«/mi»«/msubsup»«mi»#f«/mi»«mo»§nbsp;«/mo»«mi»dx«/mi»«mo»§nbsp;«/mo»«mi»y«/mi»«mo»§nbsp;«/mo»«mi»el«/mi»«mo»§nbsp;«/mo»«mi»resultado«/mi»«mo»§nbsp;«/mo»«mi»es«/mi»«mo»§nbsp;«/mo»«mi»#B«/mi»«/math»
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Feedback
Atención, esta función no es positiva en todo el intervalo. Resolviendo esta integral, el área cerrada en el intervalo donde la función es negativa está restando al área cerrada en el intervalo donde la función es positiva y no obtienes el área total.<br />#p<br />
Choice 2
Answer
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msubsup»«mo»§#8747;«/mo»«mi»#a«/mi»«mi»#b«/mi»«/msubsup»«mi»#f«/mi»«mo»§nbsp;«/mo»«mi»dx«/mi»«mo»§nbsp;«/mo»«mo»-«/mo»«mo»§nbsp;«/mo»«msubsup»«mo»§#8747;«/mo»«mi»#b«/mi»«mi»#c«/mi»«/msubsup»«mi»#f«/mi»«mo»§nbsp;«/mo»«mi»dx«/mi»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mi»y«/mi»«mo»§nbsp;«/mo»«mi»el«/mi»«mo»§nbsp;«/mo»«mi»resultado«/mi»«mo»§nbsp;«/mo»«mi»es«/mi»«mo»§nbsp;«/mo»«mi»#A«/mi»«/math»
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None
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66.66667%
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Feedback
Muy bien.<br />
Choice 3
Answer
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msubsup»«mo»§#8747;«/mo»«mi»#a«/mi»«mi»#c«/mi»«/msubsup»«mi»#f«/mi»«mo»§nbsp;«/mo»«mi»dx«/mi»«mo»§nbsp;«/mo»«mi»y«/mi»«mo»§nbsp;«/mo»«mi»el«/mi»«mo»§nbsp;«/mo»«mi»resultado«/mi»«mo»§nbsp;«/mo»«mi»es«/mi»«mo»§nbsp;«/mo»«mi»#C«/mi»«/math»
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None
100%
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83.33333%
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66.66667%
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Feedback
Atención, el área no puede ser negativa.<br />
Choice 4
Answer
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None
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Choice 5
Answer
Grade
None
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Choice 6
Answer
Grade
None
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Feedback
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Hint 2
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Created
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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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