Can someone help and express the given integral as the limit of a Riemann sum but do not evaluate: the integral from 0 to 3 of the quantity x cubed minus 6 times x, dx.

see the question I just posted below this one.

To express the given integral as the limit of a Riemann sum, we need to understand what a Riemann sum is.

A Riemann sum is an approximation of the area under a curve by dividing the interval into smaller subintervals and approximating the curve with rectangular strips. The limit of these sums as the subintervals get infinitely small approaches the exact area under the curve, which is the value of the integral.

To express the integral as a Riemann sum, we start by dividing the interval [0, 3] into n subintervals. The width of each subinterval, denoted by Δx, is given by Δx = (b - a) / n, where a and b are the endpoints of the interval.

In this case, a = 0 and b = 3, so Δx = (3 - 0) / n = 3/n.

Next, we choose a point within each subinterval, called a sample point, to evaluate the function. Let's denote the sample point in each subinterval as xi, where i ranges from 1 to n.

To approximate the integral using a Riemann sum, we sum up the areas of the rectangular strips. The area of each strip is given by the function evaluated at the sample point multiplied by the width of the subinterval:

ΔA = f(xi) * Δx

In this case, the function is given by f(x) = x^3 - 6x.

Therefore, the Riemann sum for the given integral is:

Σ [f(xi) * Δx] from i = 1 to n

To express it explicitly, we have:

(Σ [(xi^3 - 6xi) * (3/n)]) from i = 1 to n

This expression represents the Riemann sum approximation of the integral from 0 to 3 of the quantity x^3 - 6x with n subintervals.