# Help find the exact area A of region under f(x)

• Aug 27th 2006, 12:17 PM
Yogi_Bear_79
Help find the exact area A of region under f(x)
Given $\displaystyle f(x)=x^2+3$, find the exact area A of the region under f(x) on the interval [1, 3] by first computing $\displaystyle \sum_{i=1}^{n}f(x_i)\triangle x$ and then taking the limit as n → ∞.
• Aug 27th 2006, 02:33 PM
galactus
$\displaystyle {\Delta}x=\frac{3-1}{n}=\frac{2}{n}$

Right endpoint method:

$\displaystyle x_{k}=a+k{\Delta}x=1+\frac{2k}{n}$

$\displaystyle ((x_{k})^{2}+3){\Delta}x$$\displaystyle =\left(\frac{4k^{2}}{n^{2}}+\frac{4k}{n}+4\right)( \frac{2}{n})$

$\displaystyle =\frac{8(k^{2}+kn+n^{2})}{n^{3}}$

$\displaystyle \frac{8}{n^{3}}\left[\sum_{k=1}^{n}k^{2}+n\sum_{k=1}^{n}k+n^{2}\sum_{k= 1}^{n}{1}\right]$

$\displaystyle \frac{8}{n^{3}}\left[\frac{n(n+1)(2n+1)}{6}+\frac{n^{2}(n+1)}{2}+{n^{3} }\right]$

$\displaystyle \lim_{n\to\infty}\left[\frac{8}{n}+\frac{4}{3n^{2}}+44/3\right]$

Now, see what the limit is?. If you simply integrate $\displaystyle x^{2}+3$ from 1 to 3 you'll get the same result.