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anonymous 5 years ago 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

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1. amistre64

this sounds familiar ;)

2. 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?

3. 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

4. amistre64

can we double int it?

5. anonymous

i believe so

6. amistre64

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

7. amistre64

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

8. anonymous

how did you get 0 and 1 ?

9. amistre64

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

10. amistre64

thats where your graphs intersect at

11. anonymous

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

12. amistre64

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

13. amistre64

yes

14. amistre64

did i miss that up lol

15. anonymous

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

16. amistre64

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

17. amistre64

i could be wrong, trying tis in my head lol

18. anonymous

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

19. 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

20. anonymous

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

21. anonymous

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

22. amistre64

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

23. amistre64

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

24. anonymous

lol xD

25. anonymous

hmm

26. amistre64

deep breath....... ok

27. amistre64

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

28. anonymous

yup

29. amistre64

we know zero is a good one

30. anonymous

yes

31. amistre64

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

32. amistre64

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

33. amistre64

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

34. anonymous

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

35. amistre64

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

36. anonymous

xD awesome haha

37. amistre64

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

38. 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

39. amistre64

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

40. 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

41. amistre64

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

42. amistre64

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

43. amistre64

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

44. anonymous

ohhh i see

45. amistre64

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

46. anonymous

how comes its (1/7^4) ?

47. amistre64

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

48. amistre64

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

49. amistre64

dbl check it tho ;)

50. anonymous

im going over it right now :D

51. amistre64

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

52. amistre64
53. amistre64

id take the absolute value of that :)

54. anonymous

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

55. amistre64

correct; just means we got the wrong order

56. amistre64
57. anonymous

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

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