LifeIsADangerousGame
  • LifeIsADangerousGame
Help with calculus again?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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LifeIsADangerousGame
  • LifeIsADangerousGame
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TrojanPoem
  • TrojanPoem
No , you have to integrate from -1 to 2
zepdrix
  • zepdrix
Did you post the work to another problem? Hehe

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LifeIsADangerousGame
  • LifeIsADangerousGame
Yeah..I looked at the wrong problem when writing it down, I'm re-doing it
TrojanPoem
  • TrojanPoem
|dw:1449867349562:dw| When you make the dependent variable y and integrate with respect to x you get the area under the curve with respect to the x - axis and vice versa. You have to integrate with the terms of integration as the interval of the area you wanna find. If the area is separated up and under, you have to find it partially and sum.
LifeIsADangerousGame
  • LifeIsADangerousGame
So then I need to do from -1 to 0 and then from 0 to 2? and add up what I get from both of those?
TrojanPoem
  • TrojanPoem
|dw:1449867499524:dw|
TrojanPoem
  • TrojanPoem
You can , but here you don't need to. you have to do it in example like this : |dw:1449867560787:dw|
TrojanPoem
  • TrojanPoem
as you will get the first area (+ve) , second (-ve)
TrojanPoem
  • TrojanPoem
|dw:1449867614476:dw|
TrojanPoem
  • TrojanPoem
In your example: |dw:1449867727083:dw| If you want to find the area in the limit 0, -5 you have to find it in two parts. and neglect the negative sign.
LifeIsADangerousGame
  • LifeIsADangerousGame
Neglect as in leave it out? Since area can't be negative? So: \[\int\limits_{0}^{5}\] ?
TrojanPoem
  • TrojanPoem
I have mis-chosen an example. Calculate the area under the curve from -1 to -4 and tell me what you get. The answer should be 2.5 but you won't get it.|dw:1449868258260:dw| Why ? as the integration will be negative from -2 to -4 and postive from -1 to -2 so you are subtracting the areas not adding them so you have to take it from -2 to -1 (+ve) and get the area again from (-2 to -4) well, Maybe let it until you get an example of that.
mathmale
  • mathmale
I'd advoate using the simplest possible approach first, and going into further detail later if both parties are interested. Your job is to find the area under the straight line y=x+2 from x=-1 to x=2. All of this lilne is ABOVE the x-axis on the given interval. The appropriate integration formula is\[A=\int\limits_{a}^{b}f(x)dx.\]
mathmale
  • mathmale
\[Here,A=\int\limits\limits_{-1}^{2}(x+2)dx.\]
mathmale
  • mathmale
Can you do this integration? Please show your work.
LifeIsADangerousGame
  • LifeIsADangerousGame
Okay, |dw:1449868576148:dw|

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