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anonymous

  • 5 years ago

find the area bounded by y1=0, y2=-x+2, y3=sqrt(x)

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  1. amistre64
    • 5 years ago
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    y = 0 is the the x axis....let me draw a rudemintary pic of this to get an idea....

  2. amistre64
    • 5 years ago
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    is that y squared? or just the second equation of y?

  3. amistre64
    • 5 years ago
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  4. amistre64
    • 5 years ago
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    we need to determine that value of 'c' in order to piece this thing together

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  5. amistre64
    • 5 years ago
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    c equals the point where y = -x+2 and y = sqrt(x) are the same value: -x+2 = sqrt(x) x^2 -4x +4 = x^2 -4x +4 = 0 x = 1 = c

  6. amistre64
    • 5 years ago
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    \[\int\limits_{0}^{1} (-x^2+2 - \sqrt{x})dx + \int\limits_{1}^{2} (\sqrt(x)+x^2-2) dx\]

  7. amistre64
    • 5 years ago
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    \[{-x^2 \over 2} +2x -\frac{\sqrt{x}}{2}\] right? for the first one that is :)

  8. anonymous
    • 5 years ago
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    thank you very much!

  9. amistre64
    • 5 years ago
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    -1/2 - 1/2 + 2 = 1 for the area up to 'c'

  10. amistre64
    • 5 years ago
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    thats an error.... -x^2 integrates to x^3/3...so ill have to adjust for it :)

  11. amistre64
    • 5 years ago
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    -1/3 - 1/2 +2 = A

  12. amistre64
    • 5 years ago
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    2 - 3/2 = 2 - 1' 1/2 = 1/2 right?

  13. anonymous
    • 5 years ago
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    wait i dont understand how you got x^2 from the square root of x

  14. amistre64
    • 5 years ago
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    youre right; trying to do this in my head :) i put thought of (-x+2) as (-x^2 +2) when i did that :)

  15. anonymous
    • 5 years ago
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    ok

  16. amistre64
    • 5 years ago
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    [S] (-x+2 - sqrt(x)) dx ; [0,1] -x^2/2 +2x - sqrt(x)/2..... right? at 0 it equals 0 at 1 it equals 1 right?

  17. amistre64
    • 5 years ago
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    for the second run; we have the top function - bottom function sqrt(x) - (-x+2) [S] (sqrt(x) +x -2) dx ; [1,2] sqrt(x)/2 + x^2/2 -2x right?

  18. anonymous
    • 5 years ago
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    ok

  19. amistre64
    • 5 years ago
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    1/2 + 1/2 - 2 = -1 .... remember that number well need it soon enough :) sqrt(2) 4 ----- + -- - 2(2) 2 2 sqrt(2) ----- + 2 - 4 2 sqrt(2) ------ - 2 = abt. -1.29 2 -1.29 - - 1 = 1-1.29 = |-.29| = .29 as an area for the next part add the areas togher to get: 1.29; right?

  20. anonymous
    • 5 years ago
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    right

  21. amistre64
    • 5 years ago
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    now that i look at it again :) the second part is simply the area under the line (-x+2); i wonder if that changes anything :)

  22. anonymous
    • 5 years ago
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    ok :)

  23. amistre64
    • 5 years ago
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    F(x) = [S] -x+2 dx ; [1,2] F(x) = -x^2/2 + 2x F(2) = -4/2 + 4 = 2 -F(1) = -1/2 + 2 = - 1' 1/2 -------- 1/2 The area in the first part is the area under sqrt(x) from 0 to 1 F(x) = [S] sqrt(x) dx ; [0,1]

  24. amistre64
    • 5 years ago
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    F(x) = sqrt(x)/2 F(0) = 0 F(1) = 1/2 Add the areas together to get: 1/2 + 1/2 = 1 .... that sound more logical :)

  25. anonymous
    • 5 years ago
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    yes it does! thank you again!

  26. amistre64
    • 5 years ago
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    did I integrate that sqrt(x) correctly? lol...let me check. x^(1/2) x^(1+2/2) x^(3/2) 2 sqrt(x^3) ---------- = -------- = ---------- 1+2/2 3/2 3 Dont thank me yet lol.... im just an idiot in disguise till i get this right :)

  27. amistre64
    • 5 years ago
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    F(1) = 2/3..... now we can add them up :) 1 2 -- + --- 2 3 3 4 -- + --- = 7/6 for the total area :) and that my final offer 6 6

  28. anonymous
    • 5 years ago
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    haha. well you are not an idiot in disguise anymore.

  29. anonymous
    • 5 years ago
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    I got the area = 1.167 area unit. Not entirely sure though.

  30. amistre64
    • 5 years ago
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    yay!! 7/6 = 1.166.....

  31. anonymous
    • 5 years ago
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    I think the method you used is a little bit more complicated than it should be. Consider the graph you posted above, you can see that the line x=2-y is on the right of the graph of x=y^2, then the area a is: \[A=\int\limits_{0}^{1}(2-y-y^2)dy=(2y-{y^2 \over 2}-{y^3 \over 3})_{0}^{1}=2-{1 \over 2}-{1 \over 3}={7 \over 6}\]

  32. amistre64
    • 5 years ago
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    i used many methods; and eventually got smart enough to see an easier way ;)

  33. anonymous
    • 5 years ago
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    thank you very much

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