how do you find the area enclosed by a line, a parabola, and the x-axis?

- anonymous

how do you find the area enclosed by a line, a parabola, and the x-axis?

- katieb

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- anonymous

integrate :)

- anonymous

i know but integrate what

- myininaya

we need to know what parabola and what line

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## More answers

- anonymous

if you have x^2 from 0 to 1 its int (0 to 1) of x^2 dx

- anonymous

as an example

- anonymous

Find the area enclosed by \[y=\sqrt{2x}, y=6-2x, and the xaxis\]

- myininaya

first find where they intersect

- anonymous

but they intersect at three different places

- amistre64

those equation only intersect in one place :)

- anonymous

rearrange for x in the equations, your parabola will then be the lower limit, the straight line will be the upper limit and your y limits will be from 0 to the y component of the intersection point, then it is a trivial double integral

- myininaya

i will scan and show you my drawing

- myininaya

##### 1 Attachment

- anonymous

\[\int\limits_{0}^{y component of intersection}\int\limits_{(y^2)/2}^{1/2y+3}..and so on\]

- myininaya

i always have to draw picture first

- anonymous

me too by visual inspection you can simplify the problem a lot

- myininaya

dana, have you looked at the pic?

- myininaya

i also put what i integrated

- myininaya

oops that second integral should e from 2 to 3 i messed up my line a bit

- anonymous

\[\int\limits_{0}^{2}\int\limits_{.5y^2}^{-0.5y-3}dxdy\]

- anonymous

yeaa that makes sence now, thank you!

- myininaya

there was not suppose to be an e in that sentence lol

- myininaya

because the x intercept of that y=6-2x is 3 not 4

- anonymous

yea his way if fine too, the double integral is just another way you can check your answer

- myininaya

she*

- myininaya

her*

- anonymous

oh sorry

- anonymous

wasn't paying attention to the name there sorry

- anonymous

but on the other hand that double integral is pretty epic? can someone be my fan i think i deserve it for going all out on this question

- myininaya

i will be ur fan

- anonymous

thanks! i'm your fan too.

- anonymous

helped people a few times but it seems a lot of people just make an account to ask a question and then leave without saying thank you

- myininaya

your welcome dana lol

- myininaya

he didnt become our fan :(

- myininaya

dana, if you come back you could have done int(3-y/2-y^2/2,y=0..2) and got the same thing you could have look at the functions as if y-axis was x-axis

- anonymous

ohh thats true, to switch the problem around. and i did say thank you, for the record. you all were a lot of help! :)

- myininaya

lol im sorry i didn't notice im an idiot

- anonymous

are aunder a curve, use integration & subtract lower curve from upper

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