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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.

User Musa Usman
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Answer:

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

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.

User Yuvraj
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