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anonymous

  • 5 years ago

Area Under A Curve What formula's so I need/use for this worksheet ?

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    which problem exactly?

  3. anonymous
    • 5 years ago
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    all of them , i just want to know which formula's i need to use so that i can solve them all .

  4. anonymous
    • 5 years ago
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    first of all find the critical points i.e. dy/dx=0 then proceed further when the slope is +ve or -ve between those intervals

  5. anonymous
    • 5 years ago
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    r u from india?

  6. anonymous
    • 5 years ago
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    no

  7. anonymous
    • 5 years ago
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    ok...any other questions?

  8. anonymous
    • 5 years ago
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    why would you need critical points for this problem...? it's one of those rectangle approximating things that everyone hates lol

  9. amistre64
    • 5 years ago
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    everyone hates ? :)

  10. anonymous
    • 5 years ago
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    I thought critical points could help in drawing the graph...could be wrong..

  11. anonymous
    • 5 years ago
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    who needs to approximate when you can just integrate :P i always just do it and forget it

  12. amistre64
    • 5 years ago
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    id say integrate really :)

  13. amistre64
    • 5 years ago
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    why learn to walk if you are running marathons?

  14. anonymous
    • 5 years ago
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    wait...so what am i supposed to do :O

  15. amistre64
    • 5 years ago
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    you are supposed to learn that doing it the long and tedious way is proof that the shortway is better....

  16. anonymous
    • 5 years ago
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    im not running marathons im barely even walking lol

  17. amistre64
    • 5 years ago
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    when you have a function that is difficult to integrate, you can approximate the area under its curve thru numerical stuff like the trapaziod rule or simpsons rule and other time consuming methodss

  18. amistre64
    • 5 years ago
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    but these? really? lol

  19. anonymous
    • 5 years ago
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    what do you mean but these? i dont understand anything yo u just said /:

  20. amistre64
    • 5 years ago
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    [S] -x^2 +4 dx [0,2] F(x) = -x^3/3 + 4x ....at 0 its useless so solve for 2 -8/3 + 12 = -8 +36//3 = 28/3

  21. amistre64
    • 5 years ago
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    and I mess up doing simple multiplication lol....4(2) = 8 ;)

  22. amistre64
    • 5 years ago
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    -8/3 +8 = -8 +24//3 = 16/3....maybe :)

  23. anonymous
    • 5 years ago
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    integrating isnt the answer she wants though

  24. amistre64
    • 5 years ago
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    its the answer, just not the "way" :)

  25. anonymous
    • 5 years ago
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    you wont get teh same answer by approximating

  26. amistre64
    • 5 years ago
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    you will if you take the reaimann sums and sll that fun stuff...right? if I recall correclt?

  27. amistre64
    • 5 years ago
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    with inscribed and ourscribed rectangles you get the average and yadayada...

  28. anonymous
    • 5 years ago
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    yeah but reimann sum is taking an infinite number of rectangles, this is just like a few boxes

  29. amistre64
    • 5 years ago
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    then lets do boxes ;) we have to do it inscribed and outscribed.. but Im pretty sure you end up with the average being the limit of the 2..

  30. amistre64
    • 5 years ago
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    mean value theorum or some such....

  31. anonymous
    • 5 years ago
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    y'all are confusing mee /:

  32. amistre64
    • 5 years ago
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    y = -x^2 + 4 find your f(x) values for your little rectangles.... do left right sides

  33. amistre64
    • 5 years ago
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    y = -.5^2 + 4 y = -1^2 + 4 y = -1.5^2 + 4 y = -2^2 + 4

  34. amistre64
    • 5 years ago
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    the area of each "rectangle" is gonna be .5 * its y value

  35. amistre64
    • 5 years ago
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    then add the areas up

  36. amistre64
    • 5 years ago
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    I can draw a picture if you like :)

  37. anonymous
    • 5 years ago
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    yes please

  38. amistre64
    • 5 years ago
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    like this

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  39. amistre64
    • 5 years ago
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    each place a corner touches the curve; we get a value for y right?

  40. anonymous
    • 5 years ago
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    \: im still confused . i guess i got lost along y'alls "discussion" ..

  41. amistre64
    • 5 years ago
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    do you see the rectangles I drew? they have a width and a height right?

  42. anonymous
    • 5 years ago
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    yes

  43. amistre64
    • 5 years ago
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    the right edge of each rectangle hits the curve...and stops because we drew if INside the curve.... inscribed

  44. amistre64
    • 5 years ago
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    we are told that we do this every .5 units..... so the width of each rectangle is .5 makes sense?

  45. anonymous
    • 5 years ago
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    yes

  46. amistre64
    • 5 years ago
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    good, then all we need to figure out the area of any given rectangle is ...we already now its width (.5) is going to be its height...right?

  47. anonymous
    • 5 years ago
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    yep

  48. amistre64
    • 5 years ago
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    so at the first interval where x = .5; we go up until we hit the curve of y = -x+4 sine the value of "y" is our height we need to use the equation to figure out the height at x = .5 y = -.5^2 + 4 y = -2.5 +4 y = 1.5 is this right?

  49. amistre64
    • 5 years ago
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    like this

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  50. amistre64
    • 5 years ago
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    and of course I forgot how to square decimals...... -(.5*.5) = -.25 4-.25 = 3.75 thats better

  51. amistre64
    • 5 years ago
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    revised and corrected version lol

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  52. amistre64
    • 5 years ago
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    at our next step, .5 + .5 = 1. we stick x = 1 into our equation to get the height for the second rectangle; width = .5 and height =-x^2 +4. h = -(1^2) + 4 h = -1 + 4 h = 3 Area2 = .5 * 3 = 1.5

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  53. amistre64
    • 5 years ago
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    this would be easier if I didnt keep getting typos..... this is right.

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  54. amistre64
    • 5 years ago
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    at our next step, .5 + .5 +.5 = 1.5 we stick x = 1.5 into our equation to get the height for the third rectangle; width = .5 and height =-x^2 +4. h = -(1.5^2) + 4 h = -2.25 + 4 h = 1.75 Area3 = .5 *1.75 = .875

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  55. amistre64
    • 5 years ago
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    at our next step, .5 +.5 +.5+.5 = 2 we stick x = 2 into our equation to get the height for the third rectangle; width = .5 and height =-x^2 +4. h = -(2^2) + 4 h = -4 + 4 h = 0 Area4 = .5 *0 = 0 Do we really need to include this one? lets assume no :)

  56. amistre64
    • 5 years ago
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    Now we add up all the areas of our rectangles: A1 + A2 + A3 + A4 = answer 1.875 + 1.5 + .875 + 0 = answer 1.875 + 1.5 + .875 + 0 = 4.25 this is the smallest value for the area under the curve that is reasonably close to the actual area under the curve. Any of this make sense?

  57. amistre64
    • 5 years ago
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    if any of this doesnt make sense.....let me know.

  58. anonymous
    • 5 years ago
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    im sorryy if i'm making you impatient but my mind is VERY blank right now /:

  59. amistre64
    • 5 years ago
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    im not impatient dear :) just wanting to help you see what you really already know ;)

  60. amistre64
    • 5 years ago
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    tell me what you are having trouble with...and I can explain it better for you.

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