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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

Are you asking about this question (see attached file)

2. phi

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

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

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

Thanks!

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