virtus
find the area bounded by the graph y = 4x^2 1, the x axis and the ordinates x = 1/2 and x=1



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dumbcow
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\[A = \int\limits_{1/2}^{1}(4x^{2} 1) dx\]

virtus
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smartcow, i thought the ordinates were from 1/2
where did you get the 1/2 from?

dumbcow
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well when i graphed it, the parabola from 1/2 to 1/2 is below the xaxis ?

virtus
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so ....?

dumbcow
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is it bounded by the xaxis above or below ?

virtus
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dw:1328176928988:dw

virtus
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dw:1328176997411:dw

virtus
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so is the shaded the area?
dw:1328177068433:dw

dumbcow
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hmm i guess that makes more sense...why do they say it is bounded by the xaxis then

virtus
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but when i find the area i get 0 or maybe i have done some stupid error lemme check

dumbcow
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no i get 0 as well, if the limits are 1/2 to 1

dumbcow
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thats because the area under the xaxis is negative

virtus
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the answer in my solution book says the answer is 4/3 square units

dumbcow
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hmm well the area from 1/2 to 1 is 2/3
are you sure its not x=1 an x=1

virtus
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i am sure it is x=1

virtus
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perhaps the answer in the solution book is wrong then or the question is wrong

dumbcow
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dw:1328177655800:dw
\[\int\limits_{1}^{1/2}(4x^{2} 1)dx + \int\limits_{1/2}^{1}(4x^{2} 1)dx = 2/3 + 2/3 = 4/3\]
thats my best guess :)

virtus
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WOAH!!!!!!!!!!!!!!!! that is brilliant thinking
goodjob smartcow, thanks for the help greatly appreciated

dumbcow
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your welcome

FoolForMath
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But the only solution for \( 4x^21\) is \(\pm\frac 12 \)

virtus
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genius what are you trying to say?

FoolForMath
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The parabola intersects the x axis at \(\pm \frac 12 \)

FoolForMath
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This question is not correct in my opinion.

virtus
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oh i see, i think i support your opinion


virtus
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ya

FoolForMath
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ya ya :D

FoolForMath
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So I think dumcow's interpretation is not correct either.

virtus
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ya ya :D

dumbcow
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haha sorry i should have clarified , my drawing assumes the question is wrong at to get area of 4/3 the bounds must be x=1 and x=1 and xaxis

virtus
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no need to apologise! you have a case indeed hehehe

FoolForMath
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But the bound x=1 and x=1 makes no sense right?

virtus
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ya

FoolForMath
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ya ya :D

FoolForMath
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I am leaving now, take care :)

virtus
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farewell genius, and thanks for the help once again !