- anonymous

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

- chestercat

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- dumbcow

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

- anonymous

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

- dumbcow

well when i graphed it, the parabola from -1/2 to 1/2 is below the x-axis ?

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## More answers

- anonymous

so ....?

- dumbcow

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

- anonymous

|dw:1328176928988:dw|

- anonymous

|dw:1328176997411:dw|

- anonymous

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

- dumbcow

hmm i guess that makes more sense...why do they say it is bounded by the x-axis then

- anonymous

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

- dumbcow

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

- dumbcow

thats because the area under the x-axis is negative

- anonymous

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

- dumbcow

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

- anonymous

i am sure it is x=1

- anonymous

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

- dumbcow

|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 :)

- anonymous

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

- dumbcow

your welcome

- anonymous

But the only solution for \( 4x^2-1\) is \(\pm\frac 12 \)

- anonymous

genius what are you trying to say?

- anonymous

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

- anonymous

This question is not correct in my opinion.

- anonymous

oh i see, i think i support your opinion

- anonymous

See here, the area is not well defined :http://www.wolframalpha.com/input/?i=+y+%3D+4x%5E2+-1%3Bx%3D-1%2F2%3Bx%3D1%3Bx%3D0%3By%3D0

- anonymous

ya

- anonymous

ya ya :D

- anonymous

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

- anonymous

ya ya :D

- dumbcow

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

- anonymous

no need to apologise! you have a case indeed hehehe

- anonymous

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

- anonymous

ya

- anonymous

ya ya :D

- anonymous

I am leaving now, take care :)

- anonymous

farewell genius, and thanks for the help once again !

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