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byerskm2

  • 2 years ago

Find a function f for which the limit below represents the area under f on the interval [0, a]. Here we assume the rectangles used have equal widths and right endpoints as sample points. http://i49.tinypic.com/20ua7i9.jpg

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  1. satellite73
    • 2 years ago
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    any guesses for this one?

  2. satellite73
    • 2 years ago
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    no guesses? i guess \(f(x)=x^2\)

  3. satellite73
    • 2 years ago
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    on the interval \([0,1]\) divide the interval in to \(n\) parts each of length \(\frac{1}{n}\)

  4. byerskm2
    • 2 years ago
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    Just x^2? I thought it had to be more than that though

  5. satellite73
    • 2 years ago
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    the first simple point will be \(\frac{1}{n}\) the second will be \(\frac{2}{n}\) and the \(i\)th will be \(\frac{i}{n}\)

  6. satellite73
    • 2 years ago
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    then for each \(i\), \(f(\frac{i}{n})=\frac{i^2}{n^2}\)

  7. byerskm2
    • 2 years ago
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    oh wow! Dang. I hate this notation. It makes it ten times more complicated than it needs to be.

  8. satellite73
    • 2 years ago
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    multiply each by \(\frac{1}{n}\) and add the up you get \[\sum\frac{i^2}{n^3}\]

  9. satellite73
    • 2 years ago
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    well don't be hoodwinked in to thinking these are all that easy usually it is much harder to tell

  10. byerskm2
    • 2 years ago
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    I have another one a lot like this. with a sqaure root. but I'm not sure about it

  11. satellite73
    • 2 years ago
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    but the first clue should have been the \(i^2\) on the other hand if the interval had be \([1,2]\) it would have been completely different

  12. byerskm2
    • 2 years ago
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    interesting, because the i and n values would be different right?

  13. byerskm2
    • 2 years ago
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    because a-b / n ?

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