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Jaweria

I need help with one question of Calculus 1.

  • one year ago
  • one year ago

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  1. dave0616
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    what is your question?

    • one year ago
  2. Jaweria
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    My question is this: USe 5 rectangles to estimate the area under the graph of: f(x) = 2x^3+2x-5. from x=2 to x=7

    • one year ago
  3. Jaweria
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    And also tell whether your estimate is an over or an upper estimate, explain why?

    • one year ago
  4. dave0616
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    oh you are doing reimann sums? let me do some work, give me a second

    • one year ago
  5. Jaweria
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    ok thx

    • one year ago
  6. Jaweria
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    if you can check with me that I got my 5 rectangles: 35, 93, 193, -92, -128 are they right?

    • one year ago
  7. dave0616
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    hmmmmm im gonna draw up the graph so that we know we have the same thing|dw:1366346945142:dw| That is the graph that i have, roughly at least, is that the same thing that you are working with?

    • one year ago
  8. Jaweria
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    yeah

    • one year ago
  9. Jaweria
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    but can you please check my numbers plzz

    • one year ago
  10. dave0616
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    i punched this into my calculator, and the base for each rectangle would be 1, because you are using 5 rectangles, and the difference between 7 and 2 is 5. so you are basicallly looking at the height, but i got 15, 55, 131, 255, and 439. i dont see how you could get any negative because the line is going up

    • one year ago
  11. Jaweria
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    yeah i got that too but I m saying that arent we suppose to add up those rectangles?

    • one year ago
  12. Jaweria
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    so whats my answer basically?

    • one year ago
  13. Jaweria
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    Someone plzz help me

    • one year ago
  14. jojo921
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    intervals =b-a/n (7-2)/5 =1

    • one year ago
  15. dave0616
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    yea, you just add what you got for the areas, so 15+55+131+255+439=839. and it would be an under estimate because it is a left side reimann sum. meaning that the left corner of the boxes hit the line

    • one year ago
  16. Jaweria
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    i thought its over estimate because its increasing

    • one year ago
  17. dave0616
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    |dw:1366348904077:dw| that isnt to scale, but that is the basic idea of what is going on

    • one year ago
  18. Jaweria
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    oh thx

    • one year ago
  19. dave0616
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    no problem

    • one year ago
  20. Jaweria
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    one more plzz

    • one year ago
  21. dave0616
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    shoot

    • one year ago
  22. Jaweria
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    if you dont mind

    • one year ago
  23. Jaweria
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    ok

    • one year ago
  24. Jaweria
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    Use 5 rectangles to esitmate the area under the graph: y=x^3-10x^2+27x-18

    • one year ago
  25. Jaweria
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    from 1<equal to x <equalt to 6

    • one year ago
  26. Jaweria
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    |dw:1366349539104:dw|

    • one year ago
  27. dave0616
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    ok so your graph is going to look like: |dw:1366349701319:dw| roughly of course.

    • one year ago
  28. Jaweria
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    its ok

    • one year ago
  29. Jaweria
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    so how am I actually going to solve this

    • one year ago
  30. dave0616
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    are you supposed to use 5 boxes again?

    • one year ago
  31. Jaweria
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    yeah its says 5 rectangles

    • one year ago
  32. dave0616
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    does this problem have the same basis as the other problem you posted? meaning you are just looking for rectangles to find the area under the graph? ok, 5 rectangles

    • one year ago
  33. Jaweria
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    yeah

    • one year ago
  34. Jaweria
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    and it also gives you from where to where

    • one year ago
  35. Jaweria
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    which is this |dw:1366349945211:dw|

    • one year ago
  36. dave0616
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    right, so this again is the same difference as the other problem, (5) so each will have the same base length of 1

    • one year ago
  37. Jaweria
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    ok

    • one year ago
  38. Jaweria
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    can we work on this together little bit ?

    • one year ago
  39. Jaweria
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    If you dont mind?

    • one year ago
  40. dave0616
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    yes of course we can. how would you like to do that? i would like to help you in the best way that suits you

    • one year ago
  41. Jaweria
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    can you first show me how to get the value for the 1st rectangle?

    • one year ago
  42. dave0616
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    yes, i personally am using a graphing calculator because i find it is the easiest way to do it. are you using the same, or are you trying to do this by hand?

    • one year ago
  43. dave0616
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    @Jaweria

    • one year ago
  44. Jaweria
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    hand

    • one year ago
  45. Jaweria
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    we are not allow to use calculator

    • one year ago
  46. dave0616
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    ok, that is fine, that just means that we are going to have to find the value for y at every point. and our points are going to be 1,2,3,4,5,6. do you think you can solve for y using the equation for these six points? and would you like to solve them on your own, and then post them, and we can compare values?

    • one year ago
  47. Jaweria
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    ok so first I m using 1 for my y

    • one year ago
  48. dave0616
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    nope, we are gonna be solving for y, so we are going to plug 1,2,3,4,5 and 6 in for x into the equation

    • one year ago
  49. Jaweria
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    oh ok

    • one year ago
  50. dave0616
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    so solve for y, using those six values and let me know what you get

    • one year ago
  51. Jaweria
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    so let me get their values then I ll tell you

    • one year ago
  52. Jaweria
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    alright

    • one year ago
  53. dave0616
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    ok, and solve for 1,2,3,4,5 that will give us 5 rectangles

    • one year ago
  54. dave0616
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    did you get the values?

    • one year ago
  55. Jaweria
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    so far I got 10,4,0

    • one year ago
  56. dave0616
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    hmmmm, that first one isnt correct try it again

    • one year ago
  57. Jaweria
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    ok

    • one year ago
  58. Jaweria
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    its 0

    • one year ago
  59. dave0616
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    that is correct, now go ahead and fine the y value for 4 and 5

    • one year ago
  60. Jaweria
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    ok

    • one year ago
  61. dave0616
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    did you solve for them? sorry if i am rushing you, i dont mean to be

    • one year ago
  62. dave0616
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    @jaweria

    • one year ago
  63. Jaweria
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    I m sorry my stupid laptop is giving me tough time

    • one year ago
  64. dave0616
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    its all good no worries

    • one year ago
  65. Jaweria
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    I got -10

    • one year ago
  66. Jaweria
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    is that right?

    • one year ago
  67. dave0616
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    you need, two more values, but -10 is not correct, you are going to need to solve for 4 and 5. go ahead and do that and let me know what you get

    • one year ago
  68. Jaweria
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    I got -6 for 4

    • one year ago
  69. Jaweria
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    and -8 for 5

    • one year ago
  70. dave0616
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    that is correct. now we need to find the area of each of those rectangles. and the base of each of those rectangles is 1, so the area of each of those rectangles will be equal to the value, so we have 0,4,0,-6,-8. so now all we need to do is add those values.

    • one year ago
  71. dave0616
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    so A=0+4+0-6-8. what do you get for area?

    • one year ago
  72. Jaweria
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    yeah i got that too all of them

    • one year ago
  73. Jaweria
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    but then how am i suppose to find an area add them up?

    • one year ago
  74. Jaweria
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    and then divide it by some number?

    • one year ago
  75. dave0616
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    we found that the base of each of those rectangles would be 1. and the area of a rectangle is b*h. we found the height by finding the y. so in essence you are multplying each of the values we found (0,4,0,-6,-8) by one, but they will equal the values themselves. so what wer are actually doing is A=0(1)+4(1)+0(1)-6(1)-8(1)

    • one year ago
  76. dave0616
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    can you solve for area from that?

    • one year ago
  77. Jaweria
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    oh ok

    • one year ago
  78. dave0616
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    what do you get for area?

    • one year ago
  79. Jaweria
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    yeah its -10

    • one year ago
  80. Jaweria
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    I m still getting -10

    • one year ago
  81. dave0616
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    thta is right. is that what you were reffering to earlier when you said -10?

    • one year ago
  82. Jaweria
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    yup

    • one year ago
  83. dave0616
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    oh im extremely sorry, i did not realize that is what you were referrring to. i really do feel bad, im sorry. but yes it is -10, and it would still be an under estimate because we used a left corner triangle

    • one year ago
  84. Jaweria
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    its ok :)

    • one year ago
  85. Jaweria
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    so thats my answer ?

    • one year ago
  86. dave0616
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    haha ok. well im glad we went through it still. yes the area under the graph by the estimate of 5 rectangles is -10. do you understand how we came to the answer?

    • one year ago
  87. Jaweria
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    yup I completely did thanks a lot:)

    • one year ago
  88. dave0616
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    absolutely no problem!

    • one year ago
  89. Jaweria
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    I should go to sleep now its really late take care bye.

    • one year ago
  90. dave0616
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    i hear ya haha. take care and good luck in your studies

    • one year ago
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