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.

Question

Mathematics

Rey Underwood

2 April, 18:55

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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Payton Lutz · Answer 1

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.