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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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\[A = \int\limits_{1/2}^{1}(4x^{2} -1) dx\]
smartcow, i thought the ordinates were from -1/2 where did you get the 1/2 from?
well when i graphed it, the parabola from -1/2 to 1/2 is below the x-axis ?

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Other answers:

so ....?
is it bounded by the x-axis above or below ?
so is the shaded the area? |dw:1328177068433:dw|
hmm i guess that makes more sense...why do they say it is bounded by the x-axis then
but when i find the area i get 0 or maybe i have done some stupid error lemme check
no i get 0 as well, if the limits are -1/2 to 1
thats because the area under the x-axis is negative
the answer in my solution book says the answer is 4/3 square units
hmm well the area from 1/2 to 1 is 2/3 are you sure its not x=-1 an x=1
i am sure it is x=1
perhaps the answer in the solution book is wrong then or the question is wrong
|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 :)
WOAH!!!!!!!!!!!!!!!! that is brilliant thinking goodjob smartcow, thanks for the help greatly appreciated
your welcome
But the only solution for \( 4x^2-1\) is \(\pm\frac 12 \)
genius what are you trying to say?
The parabola intersects the x axis at \(\pm \frac 12 \)
This question is not correct in my opinion.
oh i see, i think i support your opinion
See here, the area is not well defined :
ya ya :D
So I think dumcow's interpretation is not correct either.
ya ya :D
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 x-axis
no need to apologise! you have a case indeed hehehe
But the bound x=-1 and x=1 makes no sense right?
ya ya :D
I am leaving now, take care :)
farewell genius, and thanks for the help once again !

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