Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width.
y=7x^2
y^2 = (1/7)x
find area of the region

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- amistre64

this sounds familiar ;)

- anonymous

helloo and yes D: i feel like i did all the steps right, but its not the right answer >< i set both the equations to eachother to find the boundaries, thats the first step right?

- anonymous

7x^2 = sqrt of (1/7)x
i square both sides to get rid of the radical so im left with
7x^4 = (1/7)x
i multiply both sides by 7, i get
49x^4 = x
subtract x from both sides, you get
49x^4 - x = 0

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- amistre64

can we double int it?

- anonymous

i believe so

- amistre64

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

- amistre64

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

- anonymous

how did you get 0 and 1 ?

- amistre64

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

- amistre64

thats where your graphs intersect at

- anonymous

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

- amistre64

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

- amistre64

yes

- amistre64

did i miss that up lol

- anonymous

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

- amistre64

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

- amistre64

i could be wrong, trying tis in my head lol

- anonymous

lol youre right, but wouldnt we have to set that second equation that y is on one side by itself, not x ?

- amistre64

y=sqrt(1/7x) ; y = 7x^2
1/7x = 49x^4
49 x^4 - 1/7x = 0
x(49x^3 - 1/7) = 0 x=0 to.....my brain shut off lol

- anonymous

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

lol xD

- anonymous

hmm

- amistre64

deep breath....... ok

- amistre64

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

- anonymous

yup

- amistre64

we know zero is a good one

- anonymous

yes

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

xD awesome haha

- amistre64

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

- amistre64

i know that this is 2 parabolas at right angles to each other; so they only meet at 2 point, to that one works and has to be it lol

- amistre64

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

- anonymous

is the 2aqrt((7x)^3)/3 the same as 2x^(3/2) divided by 21 ? cuz i put sqrt of 1/7 as (1/7)^(1/2) and found the antiderivative like that

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

ohhh i see

- amistre64

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

- anonymous

how comes its (1/7^4) ?

- amistre64

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

- amistre64

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

- amistre64

dbl check it tho ;)

- anonymous

im going over it right now :D

- amistre64

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

- amistre64

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

- amistre64

id take the absolute value of that :)

- anonymous

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

- amistre64

correct; just means we got the wrong order

- amistre64

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.