Calculus help please? :)

Use Riemann sums and limits to compute the area bounded by
f(x)=15 x^3 + 6 x
and the x axis between x=2 to x =8

Thank you!!

dx = (6/n)
x(i) = 2+(6/n)i
i = 0, n-->inf
area = sum [i = 0, n] f(x(i)) dx
= sum [i = 0, n] [15(2+(6/n)i)^3 + 6(2+(6/n)i)]

Can you finish it?


By the way, please see my comments on the other problem you asked today.

You must write the formula. Sorry if my notation looks odd. Yahoo Answers has its limitations. Know your power sums

S = ∑(i = 1 to n):f(xi)∆x
∆x = (8 - 2)/n = 6/n
xi = 2 + i∆x

S = ∑(i = 1 to n):f(2 + 6i/n)6/n
S = ∑(i = 1 to n):[15(2 + 6i/n)^3 + 6(2 + 6i/n)]6/n
S = ∑(i = 1 to n):[10(1 + 3i/n)^3 + (1 + 3i/n)]72/n
S = ∑(i = 1 to n):[10(1 + 3(3i/n) + 3(3i/n)^2 + (3i/n)^3) + (1 + 3i/n)]72/n
S = ∑(i = 1 to n):(11 + 93i/n + 270i^2/n^2 + 270i^3/n^3)72/n
S = ∑(i = 1 to n):792/n + 6696i/n^2 + 19440i^2/n^3 + 19440i^3/n^4
S = ∑(i = 1 to n):792/n + ∑(i = 1 to n):6696i/n^2 + ∑(i = 1 to n):(19440i^2/n^3 + 19440i^3/n^4)
S = 792 + 6696[n(n + 1)/2]/n^2 + 19440∑(i = 1 to n):(i^2/n^3 + i^3/n^4)
S = 792 + 3348(n^2 + n)/n^2 + 19440∑(i = 1 to n):i^2/n^3 + 19440∑(i = 1 to n):i^3/n^4
S = 792 + 3348 + 3348/n + 19440n(n + 1)(2n + 1)/(6n^3) + 19440[n(n + 1)/2]^2/n^4
S = 4140 + 3348/n + 19440(2n^3 + 3n^2 + n)/(6n^3) + 19440(n^4 + 2n^3 + n^2)/(4n^4)
S = 4140 + 3348/n + 19440(2/6 + 3/(6n) + 1/(6n^2)) + 19440(1/4 + 2/(4n) + 1/(4n^2))
S = 4140 + 3348/n + 19440/3 + 19440/(2n) + 19440/(6n^2) + 19440/4 + 19440/(2n) + 19440/(4n^2)
S = 15480 + 22788/n + 8100/n^2

As n approaches infinity,
S = 15480

Each differential element has width delta-x and height f(x*i)

x*i is = a + i*delta-x
Because as i, which is an integer, increases, i*delta-x increases by delta-x amount.

Summing all the differential elements together you get:

delta-x [Sum from i to n of f(x*i)]

delta-x = (b - a)/n

Take the limit of n to infinity, delta-x will become infinitesimally small... turning into dx.
Then you will have an integral from a to b of f(x)dx