anonymous
  • anonymous
find the area bounded by the graph y = 4x^2 -1, the x axis and the ordinates x = -1/2 and x=1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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dumbcow
  • dumbcow
\[A = \int\limits_{1/2}^{1}(4x^{2} -1) dx\]
anonymous
  • anonymous
smartcow, i thought the ordinates were from -1/2 where did you get the 1/2 from?
dumbcow
  • dumbcow
well when i graphed it, the parabola from -1/2 to 1/2 is below the x-axis ?

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anonymous
  • anonymous
so ....?
dumbcow
  • dumbcow
is it bounded by the x-axis above or below ?
anonymous
  • anonymous
|dw:1328176928988:dw|
anonymous
  • anonymous
|dw:1328176997411:dw|
anonymous
  • anonymous
so is the shaded the area? |dw:1328177068433:dw|
dumbcow
  • dumbcow
hmm i guess that makes more sense...why do they say it is bounded by the x-axis then
anonymous
  • anonymous
but when i find the area i get 0 or maybe i have done some stupid error lemme check
dumbcow
  • dumbcow
no i get 0 as well, if the limits are -1/2 to 1
dumbcow
  • dumbcow
thats because the area under the x-axis is negative
anonymous
  • anonymous
the answer in my solution book says the answer is 4/3 square units
dumbcow
  • dumbcow
hmm well the area from 1/2 to 1 is 2/3 are you sure its not x=-1 an x=1
anonymous
  • anonymous
i am sure it is x=1
anonymous
  • anonymous
perhaps the answer in the solution book is wrong then or the question is wrong
dumbcow
  • 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
  • anonymous
WOAH!!!!!!!!!!!!!!!! that is brilliant thinking goodjob smartcow, thanks for the help greatly appreciated
dumbcow
  • dumbcow
your welcome
anonymous
  • anonymous
But the only solution for \( 4x^2-1\) is \(\pm\frac 12 \)
anonymous
  • anonymous
genius what are you trying to say?
anonymous
  • anonymous
The parabola intersects the x axis at \(\pm \frac 12 \)
anonymous
  • anonymous
This question is not correct in my opinion.
anonymous
  • anonymous
oh i see, i think i support your opinion
anonymous
  • 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
  • anonymous
ya
anonymous
  • anonymous
ya ya :D
anonymous
  • anonymous
So I think dumcow's interpretation is not correct either.
anonymous
  • anonymous
ya ya :D
dumbcow
  • 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
  • anonymous
no need to apologise! you have a case indeed hehehe
anonymous
  • anonymous
But the bound x=-1 and x=1 makes no sense right?
anonymous
  • anonymous
ya
anonymous
  • anonymous
ya ya :D
anonymous
  • anonymous
I am leaving now, take care :)
anonymous
  • anonymous
farewell genius, and thanks for the help once again !

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