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.
To express the given integral as the limit of a Riemann sum, we need to partition the interval [0, 3] into subintervals and choose sample points within each subinterval to evaluate the function.
First, let's choose the number of subintervals, denoted by n. The choice of n will affect the accuracy of our approximation. For simplicity, we will divide the interval into n equal subintervals.
Next, we need to determine the width of each subinterval. Since the interval [0, 3] has a length of 3, the width of each subinterval will be 3/n.
Now, we can express the given integral as the limit of a Riemann sum:
∫[0, 3] (x^3 - 6x) dx = lim(n→∞) Σ[i=1 to n] f(xi)Δx
Where f(xi) represents the value of the function at the sample point xi within each subinterval, and Δx represents the width of each subinterval.
Substituting our function f(x) = x^3 - 6x and Δx = 3/n, we have:
lim(n→∞) Σ[i=1 to n] (xi^3 - 6xi) * (3/n)
This expression represents the limit of a Riemann sum for the given integral from 0 to 3 of (x^3 - 6x) dx without evaluating the integral.