# Can't Find My Mistake. Area between Function and its Floor Function.

• Jun 14th 2012, 06:56 AM
jml12
Can't Find My Mistake. Area between Function and its Floor Function.
Greetings everyone. For the following problem, I think I got the idea but I cannot find the mistake that I made. I know there is one because my answer is $k + \frac{5}{6}$ which is NOT
$k + \frac{11}{6}$. I am also trying to get this right before I sum each representative area between function and the floor function.

Thank you everyone.

http://i45.tinypic.com/v58jz8.png

My work

$\text{For }x\in (k,k-1)\text{ on }x\text{-axis, area = }\int_{k-1}^{k}{{{x}^{2}}+3x-1}\text{ }dx-1\cdot {{(k-1)}^{2}}+3(k-1)-1$

$= \left[ \frac{{{x}^{3}}}{3}+\frac{3}{2}{{x}^{2}}-x \right]_{k-1}^{k}-1\cdot {{(k-1)}^{2}}+3(k-1)-1$

$= \frac{{{k}^{3}}-{{k}^{3}}+3{{k}^{2}}-3k+1}{3}+\frac{3{{k}^{2}}-5{{k}^{2}}+10k-5}{2}-k+(k-1)-3\left( k-1 \right)+1$

$= k+\frac{1}{3}-\frac{5}{2}+3=k+\frac{5}{6}
$

• Jun 14th 2012, 03:11 PM
simamura
Re: Can't Find My Mistake. Area between Function and its Floor Function.
You should take interval (k,k+1) because you end up with summing up $k+\frac{11}{6}$ for k=1 to n-1.
If you take interval (k-1,k) then for k=1 this is interval (0,1) but you need to find integral from 1 to n, not from 0 to n.
So, take interval (k,k+1).
$A=\int_k^{k+1}(x^2+3x-1)dx-(k^3+3k-1)$

Also notice that your answer is $k+\frac56=(k+\frac{11}{6})-1$.
This is because you chose starting point to be 0, not 1.
Value of integral (using your formula) on interval (1-1=0,1) is $0+\frac56$, on interval (2-1=1,2) is $2+\frac56=1+\frac{11}6$ which is exactly same answer for interval (1,1+1) using required formula.
It is just a question of shifting.
• Jun 14th 2012, 03:44 PM
richard1234
Re: Can't Find My Mistake. Area between Function and its Floor Function.
A really simple solution is just to evaluate

$\int_{1}^{n} x^2 + 3x - 1 \, dx - \sum_{i=1}^{n-1} i^2 + 3i - 1$

Note that for the summation, i ranges from 1 to n-1 (because we are dealing with floor values).

$=(\frac{1}{3}n^3 + \frac{3}{2}n^2 - n) - (\frac{1}{3} + \frac{3}{2} - 1) - \sum_{i=1}^{n-1} i^2 + 3i - 1$

$= \frac{1}{3}n^3 + \frac{3}{2}n^2 - n - \frac{5}{6} - \sum_{i=1}^{n-1} i^2 + 3i - 1$.

To evaluate $\sum_{i=1}^{n-1} i^2 + 3i - 1$ we split it up to $\sum_{i=1}^{n-1} i^2 + 3 \sum_{i=1}^{n-1} i - \sum_{i=1}^{n-1} 1$

$=\frac{(n-1)(n)(2n-1)}{6} + 3\frac{n(n-1)}{2} - (n-1)$

$= \frac{1}{6} (n(n-1)(2n-1) + 9n(n-1) - 6(n-1))$ and take it on from there.
• Jun 22nd 2012, 07:43 AM
jml12
Re: Can't Find My Mistake. Area between Function and its Floor Function.
Thank youv ery much! I got it!