Area under a curve using an infinite amount of squares

[Idealistic]

Im trying to find the area under the curve y = x^3 over the interval [2,4], but im not positive whether im doing it correctly.

First, I figured the width would be 2/n because I subtracted 2 from four, the parameters of my x values. The n represents the infinite amount of rectangles that could exist.

Therefore, my points are as follows?:

x(1) = 2/n, x(2) = 4/n, x(3) = 6/n,..., x(i) = 2i/n

My heights equal x(i)^3, or 8i^3/n^3

Therefore my areas are as follows?:

A(n) = 0/n^4 + 16/n^4 + 128/n^4 + 432/n^4 + ... + ? ** here's where i'm stumped, but it could be because I made some mistakes along the way.

I know there is a common factor of 1/n^4 in each of my terms so my area expression could now be represented with:

A(n) = 1/n^4(0 + 16 + 128 + 432 + ... + ?)

However, I could factor out 16 in each of my terms and get:

A(n) = 16/n^4(0 + 1 + 8 + 27 + ... + (n(n + 1)/2)^2)

Im just fishing here I know Im sort of on the right track with this I could just be making an error of some sort.

I appreciate any help!

 

[galactus]

By using the right endpoint method, we have \(\displaystyle a+k{\Delta}x=2+\frac{2k}{n}\)

The width of each rectangle is 2/n, so we have \(\displaystyle (2+\frac{2k}{n})^{3}(\frac{2}{n})\)

When expnaded becomes \(\displaystyle \frac{16k^{3}}{n^{4}}+\frac{48k^{2}}{n^{3}}+\frac{48k}{n^{2}}+\frac{16}{n}\)

Now, here's where you use the sums of the k's. Remember the formulas?.

\(\displaystyle \frac{16}{n^{4}}\sum_{k=1}^{n}k^{3}+\frac{48}{n^{3}}\sum_{k=1}^{n}k^{2}+\frac{48}{n^{2}}\sum_{k=1}^{n}k+\frac{16}{n}\sum_{k=1}^{n}1\)

Using the formula for the sums of the squares, cubes, etc, we get:

\(\displaystyle \frac{16}{n^{4}}\frac{n^{2}(n+1)^{2}}{4}+\frac{48}{n^{3}}\frac{n(n+1)(2n+1)}{6}+\frac{48}{n^{2}}\frac{n(n+1)}{2}+16\)

This whittles down to \(\displaystyle \frac{56}{n}+\frac{12}{n^{2}}+60\)

Now, take the limit as \(\displaystyle n\to\infty\) and what do you get?.

 

[Idealistic]

I Think I was using the left endpoint method, is that why I got (2k/n)[sup:2gn2ptyv]3[/sup:2gn2ptyv] for the height and you got (2 + 2k/n)[sup:2gn2ptyv]3[/sup:2gn2ptyv] for the height?

Other than that I was pretty close i think? Can you notice anything I did wrong in my work?