Mastering The Right Riemann Sum Method: Approximating Integrals With Precision

Which of the following is a right Riemann sum for ∫831+x−−−−−√ⅆx ?

∑k=1n(4+5kn−−−−−−√⋅5n) – has the 4 + quantity and 5/n on the right side

To find a right Riemann sum for the integral ∫831+x−−−−−√ⅆx, we need to divide its interval [8, 31] into subintervals of equal width and then evaluate the function at the right endpoint of each subinterval, multiplying by the width and adding up the resulting values.

Let’s say we want to use n subintervals, so the width of each subinterval is (31 – 8) / n = 23 / n. Then, the right endpoint of each subinterval is 8 + k(23/n), where k = 1, 2, …, n.

So, the right Riemann sum for this integral with n subintervals is:

f(8 + 23/n) * (23/n) + f(8 + 2(23/n)) * (23/n) + … + f(8 + n(23/n)) * (23/n)

= √(8 + 23/n) * (23/n) + √(8 + 2(23/n)) * (23/n) + … + √(8 + n(23/n)) * (23/n)

This formula gives an approximation of the integral, which becomes more accurate as we increase the number of subintervals (i.e., as n gets larger).

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