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

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dumbcowBest ResponseYou've already chosen the best response.4
\[A = \int\limits_{1/2}^{1}(4x^{2} 1) dx\]
 2 years ago

virtusBest ResponseYou've already chosen the best response.1
smartcow, i thought the ordinates were from 1/2 where did you get the 1/2 from?
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.4
well when i graphed it, the parabola from 1/2 to 1/2 is below the xaxis ?
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.4
is it bounded by the xaxis above or below ?
 2 years ago

virtusBest ResponseYou've already chosen the best response.1
so is the shaded the area? dw:1328177068433:dw
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.4
hmm i guess that makes more sense...why do they say it is bounded by the xaxis then
 2 years ago

virtusBest ResponseYou've already chosen the best response.1
but when i find the area i get 0 or maybe i have done some stupid error lemme check
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.4
no i get 0 as well, if the limits are 1/2 to 1
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.4
thats because the area under the xaxis is negative
 2 years ago

virtusBest ResponseYou've already chosen the best response.1
the answer in my solution book says the answer is 4/3 square units
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.4
hmm well the area from 1/2 to 1 is 2/3 are you sure its not x=1 an x=1
 2 years ago

virtusBest ResponseYou've already chosen the best response.1
perhaps the answer in the solution book is wrong then or the question is wrong
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.4
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 :)
 2 years ago

virtusBest ResponseYou've already chosen the best response.1
WOAH!!!!!!!!!!!!!!!! that is brilliant thinking goodjob smartcow, thanks for the help greatly appreciated
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
But the only solution for \( 4x^21\) is \(\pm\frac 12 \)
 2 years ago

virtusBest ResponseYou've already chosen the best response.1
genius what are you trying to say?
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
The parabola intersects the x axis at \(\pm \frac 12 \)
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
This question is not correct in my opinion.
 2 years ago

virtusBest ResponseYou've already chosen the best response.1
oh i see, i think i support your opinion
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
See here, the area is not well defined :http://www.wolframalpha.com/input/?i=+y+%3D+4x%5E2+1%3Bx%3D1%2F2%3Bx%3D1%3Bx%3D0%3By%3D0
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
So I think dumcow's interpretation is not correct either.
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.4
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
 2 years ago

virtusBest ResponseYou've already chosen the best response.1
no need to apologise! you have a case indeed hehehe
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
But the bound x=1 and x=1 makes no sense right?
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
I am leaving now, take care :)
 2 years ago

virtusBest ResponseYou've already chosen the best response.1
farewell genius, and thanks for the help once again !
 2 years ago
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