anonymous
  • anonymous
find the area of the region bounded by the graphs of f(x) = x^3 + 4x^2 - 12x and g(x) = -x^2 +2x. Please show the steps. No need to explaib.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ali2x2
  • ali2x2
have you tried tigeralgebra.com?
ali2x2
  • ali2x2
http://www.tiger-algebra.com/tiger.aspx
OregonDuck
  • OregonDuck
@ali2x2 are you going to help him?

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ali2x2
  • ali2x2
@OregonDuck its ok you can help him
OregonDuck
  • OregonDuck
nevermind i was going to suggest tiger too lol
Astrophysics
  • Astrophysics
Your graph is |dw:1438787787036:dw| so your integral is \[\int\limits_{a}^{b} [g(x)-f(x)]dx\] where your interval is a=0, b = 2
Astrophysics
  • Astrophysics
\[\int\limits_{0}^{2} [(-x^2+2x)-(x^3+4x^2-12x)] dx\]
Astrophysics
  • Astrophysics
Now go ahead and integrate :-)
ali2x2
  • ali2x2
umg i thought @OregonDuck already did the work, im missing out! ;o
Astrophysics
  • Astrophysics
|dw:1438788387136:dw| a clearer graph, where f(x) is red and g(x) is blue.
Astrophysics
  • Astrophysics
So the area is |dw:1438788598493:dw| the idea for the integral is basically this \[\int\limits_{a}^{b} [ \text{Top function-Bottom function}] dx\]
Astrophysics
  • Astrophysics
@Yaros Does that make sense?
Astrophysics
  • Astrophysics
Hey, so just to make sure you got it, \[\int\limits\limits\limits_{0}^{2} [(-x^2+2x)-(x^3+4x^2-12x)] dx = \int\limits\limits_{0}^{2} (-x^3-5x^2+14x)dx\] \[[\frac{ -x^4 }{ 4 }-\frac{ 5x^3 }{ 3 }+7x^2] |_{0}^2 =-\frac{ (2)^4 }{ 4 }-\frac{ 5(2)^3 }{ 3 }+7(2)^2-0 = \frac{ 32 }{ 3 }\]

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