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anonymous

  • 5 years ago

Find the area bounded by the curves y1 = x , y2 = − x , and y3 = −x + 2

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  1. anonymous
    • 5 years ago
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    none of those are curves hehe

  2. amistre64
    • 5 years ago
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    as far as i can tell; those dont bind to each other

  3. anonymous
    • 5 years ago
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    no such area. y2 = -x and y3 = -x + 2 are parallel

  4. anonymous
    • 5 years ago
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    ooops it is sqrt of x and sqrt of -x

  5. anonymous
    • 5 years ago
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    Is y3 still -x + 2?

  6. amistre64
    • 5 years ago
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    y=sqrt(x) y=- sqrt(x) y = -x+2 will create an area

  7. amistre64
    • 5 years ago
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    there is no such thing as sqrt(-x) in real numbers lol

  8. anonymous
    • 5 years ago
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    \[y1=\sqrt{x}, y2=-\sqrt{x}, y3=-x+2\]

  9. amistre64
    • 5 years ago
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    we can flip this to its inverse as well; y = x^2; y = -x+2 and integrate from there as well

  10. amistre64
    • 5 years ago
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    x^2 = -x+2 x^2 +x - 2 = 0 x = -2, 1

  11. anonymous
    • 5 years ago
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    of course there is \[\sqrt{-x}\] domain is non-positive numbers!

  12. amistre64
    • 5 years ago
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    the area of -x+2 from -2 to 0 = [S] [-x+2] - [x^2] dx ; [-2,0]

  13. amistre64
    • 5 years ago
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    sqrt(-x) is complex imaginary numbers....

  14. amistre64
    • 5 years ago
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    -x^2/2 +2x - x^3/3 ; [-2,1]

  15. anonymous
    • 5 years ago
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    wait... how did you get x^2 if you squared the \[\sqrt{x}\]

  16. anonymous
    • 5 years ago
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    do you know how to invert a function?\[f \rightarrow f^{-1}\]

  17. amistre64
    • 5 years ago
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    -1/2 + 2 - 1/3 is one part of it: 7/6 -4/2 - 4 + 8/3 = -10/3 -10/3 - 7/6 = -27/6 area = 27/6....if i did it right :)

  18. amistre64
    • 5 years ago
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    y = sqrt(x) and y = -sqrt(x) inverts to y = x^2 :)

  19. anonymous
    • 5 years ago
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    oh ok

  20. amistre64
    • 5 years ago
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    y = -x+2; ionverts to y = -x+2 lol

  21. anonymous
    • 5 years ago
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    i disagree. no one says -x is negative. In any cases if you draw the picture and then rotate 90 degrees clockwise you see \[y=x^2\] and \[y=x+2\] they intersect at -1 and 2 so integrate \[\int^2_{-1} x+2-x^2\,dx\]

  22. anonymous
    • 5 years ago
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    ok

  23. anonymous
    • 5 years ago
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    i got that the 3 bound are 0, 1, 4

  24. anonymous
    • 5 years ago
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    You want the intersection of \[x^2\] and \[y=x+2\] \[x^2=x+2\] \[x^2-x-2=0\] \[(x-2)(x+1)=0\] \[x=-1, x=2\]

  25. anonymous
    • 5 years ago
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    but none of the curves touch at -1.

  26. amistre64
    • 5 years ago
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    the intersection of x^2 and -x+2 was the original

  27. amistre64
    • 5 years ago
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    its odd that you disagree that sqrt(-x) is not a real number.....

  28. amistre64
    • 5 years ago
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    lovhap; did you get a right answer :)

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