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I am having some trouble with this practice problem that is:

Let $f(x) = 2x$ on $[0,3]$. Calculate the upper and lower Riemann sums with respect to the partition $P = \{0,1,\frac{3}{2} , 2,3\}$

The trouble that I am having is that we have usually done this kind of example for an arbitrary partition and then we find the k-th partition and use $[x_{k-1},x_{k}]$ for the upper and lower sums to then show that they agree but I am kinda uncertain of how to go about this problem with the specified partition. Any suggestions would be greatly appreciated.

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    $\begingroup$ $2x$ is increasing on $[0,3]$ so upper sum goes with right enpoints, lower sum with left endpoints. $\endgroup$ – coffeemath Apr 17 '17 at 15:01
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Your partition allows you to cut $[0,3]$ in sub intervals. As your function is increasing, it takes its lowest value on an interval at the left end point and the highest at the right hand point. Then you apply your formula, and if I didn't do a mistake you get something like:

The lower sum is $s=0\times (1-0)+2\times (3/2-1) + 3 \times (2-3/2) + 4\times (3-2)=13/2$,

and the upper sum is $S=2\times (1-0)+3\times (3/2-1) + 4 \times (2-3/2) + 6\times (3-2)=23/2$.

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Your partition allows you to cut [0,3] in sub intervals. As your function is increasing, it takes its lowest value on an interval at the left end point and the highest at the right hand point. Then you apply your formula, and if I didn't do a mistake you get something like:

The lower sum is s=0×(1−0)+2×(3/2−1)+3×(2−3/2)+4×(3−2)=13/2,

and the upper sum is S=2×(1−0)+3×(3/2−1)+4×(2−3/2)+6×(3−2)=23/2

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