Finding the lower and upper sums of $f(x)=2x+1$ I'm trying to find the upper and lower sums for $f(x)=2x+1$ on the interval $[1,4]$.
$P_6$ is the partition of $[1,4]$ consisting of 7 equally space points, $1, \frac{3}{2}, 2, \frac{5}{2},3,\frac{7}{2},4$
So my goal is to find $U(f,P_6)$ and $L(f,P_6)$
Without entering all of my steps I found $U(f,P_6)$ using the following:
I used the fact that $\Delta x=\frac{3}{n}$ and $f(x_i)=2\left(\frac{3i}{n}\right)+1$
When I put them together I get:
$U(f,P_6)=\sum_{i=1}^{n}\left(2\left(\frac{3i}{n}\right)+1\right)\left(\frac{3}{n}\right)=\frac{3(8n+6)}{2n}$
And for the lower sum I put together the following:
$L(f,P_6)\sum_{i=1}^{n}\left(2\left(\frac{3i}{n}-\frac{3}{n}\right)+1\right)\left(\frac{3}{n}\right)=\frac{3(4n-3)}{n}$
Did I go about this the right way?

$f(x)=2x+1$ which is increasing function on $[1,4]$ and 
$P_6={1,\frac{3}{2},2,\frac{5}{2},3,\frac{7}{2},4}$ are subdivision.
$I_1=[1,\frac{3}{2}],I_2=[\frac{3}{2},2],I_3=[2,\frac{5}{2}],I_4=[\frac{5}{2},3],I_5=[3,\frac{7}{2}],I_6=[\frac{7}{2},4]$
each interval has length $\frac{1}{2}$
 So,
$|I_k|=\frac{1}{2} \forall k=1,2,\dots,6 $ 
Now using the subdivision of $[1,4]$. calculate $m[f,I_k]$  and $M[f,I_k]$
then use 
Now calculate 
$m[f;I_1]=m[f;1]=2(1)+1=3 $         ;$  M[f;I_1]=M[f;\frac{3}{2} ]=2(\frac{3}{2})+1=4 $
$m[f;I_2]=m[f;\frac{3}{2}]=2(\frac{3}{2})+1=4$  ;
$M[f;I_2]=M[f;2]=2(2)+1=5$
$m[f;I_3]=m[f;2]=2(2)+1=5$ ;
$M[f;I_3]=M[f;\frac{5}{2}]=2(\frac{5}{2})+1=6$
$m[f;I_4]=m[f;\frac{5}{2}]=2(\frac{5}{2})+1=6$;
$M[f;I_4]=M[f;3]=2(3)+1=7$
$m[f;I_5]=m[f;3]=2(3)+1=7$;
$M[f;I_5]=M[f;\frac{7}{2}]=2(\frac{7}{2})+1=8$
$m[f;I_6]=m[f;\frac{7}{2}]=2(\frac{7}{2})+1=8$
$M[f;I_6]=M[f;4]=2(4)+1=9$
$L[f,P_6]=\sum_{k=1}^{6}m[f,I_k]|I_k|=\frac{1}{2}\{3+4+5+6+7+8\}=\frac{33}{2}$
$U[f,P_6]=\sum_{k=1}^{6}M[f,I_k]|I_k|=\frac{1}{2}\{4+5+6+7+8+9\}=\frac{39}{2}$
