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MoonlitFate

  • one year ago

How to find the area under a curve?

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  1. modphysnoob
    • one year ago
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    you integrate

  2. electrokid
    • one year ago
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    @modphysnoob haha thats no help

  3. MoonlitFate
    • one year ago
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    *Use the left and right endpoints and the given number of rectangles to find two approximations of the area of the region between the graph of the function and the x-axis over the given interval. g(x) = 2x^2-x-1, [2,5], 6 rectangles

  4. goformit100
    • one year ago
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    Use To Integrate

  5. modphysnoob
    • one year ago
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    this is approximation

  6. modphysnoob
    • one year ago
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    I will draw a picture and explain this

  7. modphysnoob
    • one year ago
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    |dw:1364087469983:dw| this is left hand approximation

  8. modphysnoob
    • one year ago
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    |dw:1364087569386:dw| this is right hand approximation

  9. modphysnoob
    • one year ago
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    see the difference?

  10. MoonlitFate
    • one year ago
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    One is above the curve, and one is below it, to put it simply. :)

  11. modphysnoob
    • one year ago
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    right, but to be more precise in left hand approximation, the left side touch the curves, in right hand approx, the right side does

  12. MoonlitFate
    • one year ago
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    You mean that all of the left endpoints of each rectangle touch the curve for the right approximation and the same for the left? Just making sure that I follow.

  13. MoonlitFate
    • one year ago
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    Err, left approximation, not right in that first sentence.

  14. modphysnoob
    • one year ago
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    yes, right, so as you said, in left hand over approximate, right hand under approximate

  15. MoonlitFate
    • one year ago
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    All right. I know that you have divide the interval into subintervals. Since it's 6 rectangles, that means 6 subintervals, right?

  16. modphysnoob
    • one year ago
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    right,

  17. modphysnoob
    • one year ago
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    so from 2 to 5 5- 2 ------------------- 6 3/6 1/2

  18. modphysnoob
    • one year ago
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    so each rectangle is 1/2 wide

  19. MoonlitFate
    • one year ago
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    Yeah, I was getting there! So, I understand that part. :)

  20. modphysnoob
    • one year ago
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    |dw:1364088406811:dw|

  21. MoonlitFate
    • one year ago
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    So, for left hand approximation, I would use the left end points for each subinterval. So, the length of each rectangle would be g(x) evaluated at each left endpoint of that subinterval?

  22. modphysnoob
    • one year ago
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    yes

  23. MoonlitFate
    • one year ago
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    Okay. This is starting to make a little bit of sense, :) One big thing that confuses me is the Sigma notation that is adding in with this. :/

  24. modphysnoob
    • one year ago
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    sigma just mean sum |dw:1364090915002:dw|

  25. MoonlitFate
    • one year ago
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    Yes, I know that much. It's just figuring out the pattern is where I have trouble.

  26. MoonlitFate
    • one year ago
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    I mean, I know that's going to be all the areas of rectangles added together.

  27. modphysnoob
    • one year ago
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    yep, that's all there is to it

  28. MoonlitFate
    • one year ago
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    I'll see if I can find some videos to help, too. :) Easier to grasp things when you see examples being worked out. But, at least, I understand a bit more! I can do this!

  29. modphysnoob
    • one year ago
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    I assume you are in calculus 2

  30. MoonlitFate
    • one year ago
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    Calculus I.

  31. modphysnoob
    • one year ago
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    if you want I can do example for you

  32. MoonlitFate
    • one year ago
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    @modphysnoob - I know this a late reply, but by all means go for it, if you don't mind. :)

  33. modphysnoob
    • one year ago
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    so let's do a parabola f(x)= x^2 |dw:1364157776078:dw|

  34. modphysnoob
    • one year ago
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    |dw:1364157838998:dw|

  35. modphysnoob
    • one year ago
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    so we wanna approximate the area under the curve using 2 rectangle so first rectangle would be from 0 to 1 1 to 2

  36. MoonlitFate
    • one year ago
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    Okay.

  37. modphysnoob
    • one year ago
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    0 1 2 in left approximation we will take 2 rectangles with height at 0 and 1 and multiply by width =1 f(0)*1+f(1)*1 in right approoximation , we take height at 1 and 2 f(1)*1+f(2)*1

  38. MoonlitFate
    • one year ago
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    Okay, I follow you so far. :) Sorry for the delay in answering; internet problems.

  39. modphysnoob
    • one year ago
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    well, there you have it , left and right approximation

  40. modphysnoob
    • one year ago
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    you will learn trapizoid approximation which is average between left and right hand

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