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 3 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
 3 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

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

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

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

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

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

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

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

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

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.4thats because the area under the xaxis is negative

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

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

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

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.4dw: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 :)

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

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

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

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

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

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

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.0See 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

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

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.4haha 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

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

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

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

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