Integration (1 Viewer)

artosis

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Strangely, the answer is false.

Since the function is even, shouldn't the answer be true?

 
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SpiralFlex

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Sketch the graph. For the second integral form you can't take the lower boundary as . Since there are vertical asymptotes at and . Hence the answer is false.
 
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Omnipotence

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Yeah, test whether it is even or not. In this case, it is even, so the rule satisfies.
 

artosis

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Yeah, test whether it is even or not. In this case, it is even, so the rule satisfies.
That's what I thought. But the rule doesn't satisfy this question.
SpiralFlex seemed to have gotten the answer. Am I really suppose to sketch a graph each time before changing the limits?

wierd question, do you mind posting the solution
Done.
 

SpiralFlex

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Hi?

That's what I thought. But the rule doesn't satisfy this question.
SpiralFlex seemed to have gotten the answer. Am I really suppose to sketch a graph each time before changing the limits?



Done.
Graphically is the only safest method I know or maybe a HSCer will tell you another method.

What about this?





For real values of ,

Hence,





, [Since , ]

Hence the boundaries of the integral does not satisfy the domain.
 
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khfreakau

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Spiral is correct. Keep in mind the even function rule only applies if the function is CONTINUOUS across all values within limits. You can't say undefined = undefined, there's no such thing :)
 

SpiralFlex

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Spiral is correct. Keep in mind the even function rule only applies if the function is CONTINUOUS across all values within limits. You can't say undefined = undefined, there's no such thing :)
Can you also check my method on top? I am not sure if it's theoretically correct. I just thought of it on the spot.
 

xV1P3R

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I think it would be enough if you said:

4x² - 1 > 0
x² > 0.25
x > 0.5, x < -0.5
Curve is undefined for x between 0.5 and -0.5, therefore integral doesn't exist

ie. pretty much what the answer says.

@Spiralflex, you're making it more complicated than it should be.
 

SpiralFlex

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I think it would be enough if you said:

4x² - 1 > 0
x² > 0.25
x > 0.5, x < -0.5
Curve is undefined for x between 0.5 and -0.5, therefore integral doesn't exist

ie. pretty much what the answer says.

@Spiralflex, you're making it more complicated than it should be.
Okay. Thanks, I will use your method once I reach Extension 2. :)
 

SpiralFlex

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Oh damn. Do I sense a hint of "OH SNAPPERY" sarcasm?
I don't get it? No I was being serious. I didn't see that method. Viper saved me time and potential marks in the HSC. Hence he was repped.
 

K4M1N3

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Oh lol :S it sounded like you were being a smart ass cos you could answer an Ext2 question without doing ext2.....but nvm.
 

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