NEED Calculus Help!! urgent?

Question

For the intergral -(x^2)/3 +6 with interval [0,3]

Rewrite the integral as a function of n without any summation sign using the right hand endpoints.

plz help

Answer 1

It looks like you mean the Riemann sum rather than the integral. Otherwise there is no n.
So assuming that we use n equally-spaced intervals, and using the right-hand endpoints as specified, we have
x_i = 3i/n, i = 1, ..., n
f(x_i) = -(x_i)^2/3 + 6
= -(3i/n)^2/3 + 6
= -3i^2/n^2 + 6
The Riemann sum is Σ(i=1 to n) f(x_i)Δx_i
= Σ(i=1 to n) [-3i^2/n^2 + 6][3/n]
= Σ(i=1 to n) [-9i^2/n^3 + 18/n]
= -9/n^3 Σ(i=1 to n) [i^2] + 18/n (n-1+1)
= 18 - 9/n^3 [n(n+1)(2n+1) / 6]
= 18 - (3/2) (n+1)(2n+1) / n^2
or, if you prefer,
= 18 - (3/2) (2 + 3/n + 1/n^2)
= 15 - 9/(2n) - 3/(2n^2).

It's clear that as n->∞, this will go to 15.

Answer 2

http://inlinethumb44.webshots.com/2987/2666639370100400651S600x600Q85.jpg

hey, the answer is on the above link. its an image.
go check it out.

u might mean that u want the riemann sum. if it is so, scarlet's answer is fine, else u don't get any 'n' in definite integral.

Answer 3

can't help with it...but I do know of a site that may help

webmath.com

give it a try and good luck