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Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your sample points.)

∫6 to 3 x / (2 + x^5) dx
lim n→[infinity] Σn to i = 1 _____________.

User Manita
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The limit of the Riemann sum is ∫6 to 3 x / (2 + x^5) dx ≈ lim n→∞ Σ(i=1 to n) [ (6 + 6i/n) / (2 + (6 + 6i/n)^5) ] * (-3/n).

Here is the integral expressed as a limit of Riemann sums using right endpoints as sample points:

∫6 to 3 x / (2 + x^5) dx ≈ lim n→∞ Σ(i=1 to n) [ (6 + 6i/n) / (2 + (6 + 6i/n)^5) ] * (6/n)

Subintervals: We subdivide the interval [6, 3] into n subintervals of equal width.

Each subinterval has width Δx = (b - a) / n = (3 - 6) / n = -3/n.

Right endpoints: We choose the right endpoint of each subinterval as the sample point.

For the ith subinterval, the right endpoint is xi = a + iΔx = 6 + i(-3/n).

Function values: We evaluate the function f(x) = x / (2 + x^5) at each sample point xi.

Riemann sum: We form the Riemann sum by multiplying the function value at each sample point by the width of the subinterval and summing over all subintervals.

Limit: We take the limit of the Riemann sum as the number of subintervals n approaches infinity. This limit represents the definite integral.

Therefore, the limit of the Riemann sum is:

∫6 to 3 x / (2 + x^5) dx ≈ lim n→∞ Σ(i=1 to n) [ (6 + 6i/n) / (2 + (6 + 6i/n)^5) ] * (-3/n)

User Csharpbd
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