# Calculus I Help?

• Jul 19th 2006, 11:31 AM
Nimmy
Calculus I Help?
Find the Riemman Sum of the integration of x^3 from 1 to 2.
• Jul 19th 2006, 12:23 PM
ThePerfectHacker
Quote:

Originally Posted by Nimmy
Find the Riemman Sum of the integration of x^3 from 1 to 2.

It depends how you want to partion the region.
I will partition the region with right-endpoint.

Take interval $[1,2]$ divide it in $n$ equal regions. Thus, each one has length $\Delta x=\frac{2-1}{n}=\frac{1}{n}$.
Thus, the Riemann sum is,
$\sum_{k=1}^n f(a+\Delta x\cdot k)\Delta x$
Rewrite is as,
$\sum_{k=1}^n f\left(1+\frac{k}{n} \right) \frac{1}{n}$
Thus,
$\sum_{k=1}^n \left( \frac{n+k}{n} \right)^3\cdot \frac{1}{n}$
Which simplifies to,
$\sum_{k=1}^n \frac{(n+k)^3}{n^4}$

Do you also want to know its value?
• Jul 19th 2006, 12:33 PM
galactus
Do you mean the tedious technical approach?.

Using right endpoint.

$y=x^{3}\;\text{over the interval}\;\ [1,2]$

${\Delta}x=\frac{2-1}{n}=\frac{1}{n}$

$x_{k}=a+k{\Delta}x=1+k{\Delta}x=1+\frac{k}{n}$

$\left(1+\frac{k}{n}\right)^{3}{\Delta}x=(1+\frac{k }{n})^{3}(\frac{1}{n})$

Expanding and summing:

$\sum_{k=1}^{n}\left(\frac{k^{3}}{n^{4}}+\frac{3k^{ 2}}{n^{3}}+\frac{3k}{n^{2}}+\frac{1}{n}\right)$

Now, remember the 'identities':

$k^{3}=\frac{n^{2}(n+1)^{2}}{4}$

$k^{2}=\frac{n(n+1)(2n+1)}{6}$

$k=\frac{n(n+1)}{2}$

Now, use those and it all whittles down to:

$\lim_{n\to\infty}\frac{15n^2+14n+3}{4n^{2}}=\frac{ 15}{2}$

Perform the 'easy' integration and see if you get the same thing.