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tavale21 Group Title

approximate the area under each curve over the specified interval by using the indicated number of subintervals (or rectangles) and evaluating the function at the right-hand endpoints of the subintervals. f(x)=x^3 from x=0 to x=3; 3 subintervals

  • one year ago
  • one year ago

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  1. wio Group Title
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    |dw:1353031457071:dw|

    • one year ago
  2. wio Group Title
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    You basically want to find the area of the three rectangles. Make sense?

    • one year ago
  3. tavale21 Group Title
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    sort of? how would i find the area's ?

    • one year ago
  4. wio Group Title
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    Well, for one, what is the area of a rectangle?

    • one year ago
  5. tavale21 Group Title
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    is it from 0 to 3?

    • one year ago
  6. wio Group Title
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    I'm talking about in the general sense

    • one year ago
  7. tavale21 Group Title
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    that's all that the book has. I looked at another problem and it is f(x) = 4x-x^2 from x=0 to x=2; 2 subintervals and the answer was 7 square units. how did they get that?

    • one year ago
  8. wio Group Title
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    Okay, I just wanted to say, the area of a rectangle is length times width.

    • one year ago
  9. wio Group Title
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    So our first task is to find the width of the rectangles.

    • one year ago
  10. wio Group Title
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    We will split up the interval equally into sub intervals... So we know with n subintervals on the interval [a,b], the width w is:\[ nw = b-a \implies w=\frac{b-a}{n} \]In this case, b = 3, a =0, and n = 3

    • one year ago
  11. wio Group Title
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    So what is the width of each rectangle?

    • one year ago
  12. tavale21 Group Title
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    ok

    • one year ago
  13. tavale21 Group Title
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    3

    • one year ago
  14. wio Group Title
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    That is the width of the entire interval, not the width of each subinterval/rectangle

    • one year ago
  15. wio Group Title
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    \[ \frac{3-0}{3} = 1 \]

    • one year ago
  16. tavale21 Group Title
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    oh ok. I see now b-a/n=1

    • one year ago
  17. wio Group Title
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    Now, to find the length of the interval is a bit tricky.

    • one year ago
  18. wio Group Title
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    We need to pick some x in each subinterval. The length is then f(x).

    • one year ago
  19. wio Group Title
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    But which x do we pick? It depends on the problem. This problem says that you want to pick the right-hand endpoints, so we want to pick the x such that x is the highest value of x for that subinterval.

    • one year ago
  20. wio Group Title
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    The first interval is from a to a+w, in this case 0 to 1. Since 1 is highest, right hand side, we use f(1) for the length of the interval.

    • one year ago
  21. wio Group Title
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    So the first interval is \(1*f(1) = 1 * 1^3 = 1\)

    • one year ago
  22. wio Group Title
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    The second interval is between a+w and (a+w)+w

    • one year ago
  23. wio Group Title
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    So the highest point is a+w+w = 0+1+1 = 2

    • one year ago
  24. wio Group Title
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    So the second subinterval area is \(1*f(2) = 1*2^3=1*8=8\)

    • one year ago
  25. wio Group Title
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    Do you think you can do the third sub interval?

    • one year ago
  26. tavale21 Group Title
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    can you tell me the steps sorry im not good at this?

    • one year ago
  27. wio Group Title
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    |dw:1353032610122:dw|

    • one year ago
  28. wio Group Title
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    Okay, first step is to find with width of the rectangles. Next step is to find the length of each rectangle.

    • one year ago
  29. wio Group Title
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    Then you add the areas of all the rectangles. Do you at least understand the high level idea of what we are doing?

    • one year ago
  30. tavale21 Group Title
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    are we adding the last f(3) to 8

    • one year ago
  31. tavale21 Group Title
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    the answer f(3) * 8

    • one year ago
  32. wio Group Title
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    Nope. All I did was show that rectangle 1 has area 1, and rectangle 2 has area 8

    • one year ago
  33. wio Group Title
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    So you need to find width times f(3) to find the area of rectangle 3. Then you add up the areas of all rectangles.

    • one year ago
  34. tavale21 Group Title
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    1*3^3=27

    • one year ago
  35. tavale21 Group Title
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    27+8+1= 36

    • one year ago
  36. tavale21 Group Title
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    can you explain how the left hand works? to the same problem? it makes sense now? thank you!!!

    • one year ago
  37. wio Group Title
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    |dw:1353033340267:dw|

    • one year ago
  38. wio Group Title
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    For the right hand, we use f(a+w), f(a+2w), f(a+3w), ... , f(a+(n-1)w), f(a+nw) For the left hand, we use f(a), f(a+w), f(a+1w), ... , f(a+(n-2)w), f(a+(n-1)w)

    • one year ago
  39. tavale21 Group Title
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    ok

    • one year ago
  40. wio Group Title
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    How do you understand how I'm getting them?

    • one year ago
  41. tavale21 Group Title
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    from the picture you drawn up you use the rectangles starting from the left to the right.

    • one year ago
  42. wio Group Title
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    You will learn later, a method to do this with infinite subintervals. It's called a definite integral.

    • one year ago
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