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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 ?

so ....?

is it bounded by the x-axis above or below ?

|dw:1328176928988:dw|

|dw:1328176997411:dw|

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

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

ya

ya ya :D

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

ya ya :D

no need to apologise! you have a case indeed hehehe

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

ya

ya ya :D

I am leaving now, take care :)

farewell genius, and thanks for the help once again !