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7 January, 14:23

We start by subdividing [0,5] into n equal width subintervals [x0, x1],[x1, x2], ...,[xn-1, xn] each of width Δx. Express the width of each subinterval Δx in terms of the number of subintervals n.

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  1. 7 January, 14:50
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    Therefore, Δx=5/n, when have n intervals.

    Step-by-step explanation:

    From exercise we have interval [0,5]. So the length of the given interval is 5-0=5. Since all intervals [x0, x1],[x1, x2], ...,[xn-1, xn] are equal in width.

    We know that their width is Δx. We conclude that width of each subinterval Δx in terms of the number of subintervals n is equal 5/n.

    Therefore, Δx=5/n, when have n intervals.
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