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)
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?
lim as x-> infinity of sigma from i=2 to n of ((2+3i/n^2+(2+3i/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 + iĪx = 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(nāā) Ī£(i=1 to n) [f(x_i) * Īx]
= lim(nāā) Ī£(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(nāā) Ī£(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.