find an expression for the area under the graph of f(x)= (x^2)+x from x=2 to x=5 as a limit of a riemann sum (do not need to evaluate).

the answer i got was:
lim as x-> infinity of sigma from i=2 to n of (2+3i/n)^2+(3i/n)(3/n)

is this correct?

sorry, I typed the answer wrong. Can you please check this answer?

lim as x-> infinity of sigma from i=2 to n of ((2+3i/n^2+(2+3i/n))(3/n)

To find the expression for the area under the graph of the function f(x) = x^2 + x from x = 2 to x = 5 as a limit of a Riemann sum, we can divide the interval [2, 5] into n subintervals of equal width. The width of each subinterval will be Δx = (5 - 2) / n = 3/n.

Now, let's choose sample points for each subinterval. We can use the right endpoint of each subinterval. So, for the ith subinterval, the sample point will be x_i = 2 + iΔx = 2 + i(3/n) = 2 + (3i/n).

Now, let's focus on finding the height of each rectangle in the Riemann sum. The height of each rectangle is given by the function f evaluated at the sample point. So, the height of the ith rectangle will be f(x_i) = (x_i)^2 + x_i = ((2 + 3i/n)^2) + (2 + 3i/n).

Finally, we can write the expression for the area under the graph of f(x) as a limit of a Riemann sum using the sigma notation:

Area = lim(n→∞) Σ(i=1 to n) [f(x_i) * Δx]
= lim(n→∞) Σ(i=1 to n) [((2 + 3i/n)^2) + (2 + 3i/n)] * (3/n)

Therefore, your expression for the area under the graph of f(x) as a limit of a Riemann sum is:

lim(n→∞) Σ(i=1 to n) [((2 + 3i/n)^2) + (2 + 3i/n)] * (3/n)

Please note that this expression represents the limit as the number of subintervals approaches infinity, and it does not need to be evaluated at a specific value.