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this sounds familiar ;)

can we double int it?

i believe so

y = sqrt(1/7x) y=7x^2
{SS} dy dx ; x=[0,1]; y = [7x^2,sqrt(1/7x)] right?

{S} sqrt(1/7x) - 7x^2 dx ; [0,1]

how did you get 0 and 1 ?

sqrt(1/7) 2x sqrt(x/3) - 7/x^3/3

thats where your graphs intersect at

to find the intersection, you set the equations equal to eachother right?

2sqrt(1/21)-7/3 i believe is the area

yes

did i miss that up lol

thats what i did at first, but i got x = 0 and x = cube root of 1/49

y= 7x^2; and x=7y^2

i could be wrong, trying tis in my head lol

1/7x(343x^3 - 1) = 0

x = 0 and x = cube root of 1/343 ?

wolfram tells me that (1,1/7) is the solution to these; lets try that out :)

sqrt(1/7*1) = sqrt(1/7)
7*1 = 7....... ack!!

lol xD

hmm

deep breath....... ok

y=7x^2 and y^2 = 1/7x right?

yup

we know zero is a good one

yes

y =1 1 7
---- = 1/7; at x=1/7, y = 1/7 ; thats y=x^2
7 7

at x = 1/7; we get y = sqrt(1/7 1/7) = sqrt(1/49) = 1/7

so lets int this thing from 0 to 1/7 :)

im sorry im a little lost, how did you get 1/7 for x ?

[S] 7x^2 - sqrt(1/7x) dx ; [0,1/7] ....i guessed and was right :)

xD awesome haha

i had an inclination, but couldnt verify it till i just plugged it in ;)

7x^3/3 - 2sqrt((7x)^3)/3 ??

sqrt(1/7)* sqrt(x) = sqrt(1/7)* 2sqrt(x^3)/3
7x^3/3 - 2sqrt(1/7x^3)/3 right?

the sqrt(1/7) is actually a constant, so we can pull it out of the way and just do sqrt(x)

sqrt(ab) = sqrt(a) * sqrt(b)

ohhh i see

1/49(3) - 2sqrt(1/7^4)/3 should be the area

how comes its (1/7^4) ?

1 2sqrt(1/2401)
--- - ------------
147 3
(1/7) * (1/7)^3 = 1/7^4

as long as i integrated it right, and im pretty sure i did :)

dbl check it tho ;)

im going over it right now :D

\[\int\limits_{} (7x^2 - \sqrt{\frac{1}{7}x} )dx\]

http://www.wolframalpha.com/input/?i=int%287x^2+-sqrt%28x%2F7%29%29dx++from+0+to+1%2F7

id take the absolute value of that :)

thats what i got too :) and i jus ttake the absolute value of that since its negative right ?

correct; just means we got the wrong order

see...http://www.wolframalpha.com/input/?i=int%28sqrt%28x%2F7%29-7x^2%29dx++from+0+to+1%2F7

gotcha. thank you once again. always coming to my rescue :D haha youre the best !