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anonymous

  • one year ago

I'm on problem set 6, question 3B-5. Why is the integrand in the solution sin(bx) as opposed to sin(x)?

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  1. phi
    • one year ago
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    Are you asking about this question (see attached file)

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  2. phi
    • one year ago
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    The Riemann sum is computed by dividing the area under the curve into thin rectangles and adding up the area of these rectangles. say we have "n" rectangles |dw:1437134049812:dw|

  3. phi
    • one year ago
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    the height of the kth rectangle is f( x), or in this case sin( b x) where x = k * width of each rectangle. if we have n rectangles over the interval from x=0 to x=1, each rectangle has width 1/n, and the kth rectangle will have area \[ \sin(b⋅k⋅width)⋅width \\ \sin\left(b⋅k⋅\frac{1}{n}\right)⋅\frac{1}{n}\]

  4. phi
    • one year ago
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    notice that we are working with sin(b x) Just sin(x) would mean the kth rectangle has area \[ \sin\left( k \cdot \frac{1}{n}\right) \cdot \frac{1}{n}\]

  5. anonymous
    • one year ago
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    Thanks!

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