# Midpoints and Riemann sums

I am trying to figure out these questions and I think I understand how this complicated formula works. I do not have it memorized but I do have it written down for reference.

If $f(x)=e^x-2$, $0\le x\le2$ then find the Riemann sum with $n=4$ correct to six decimal places, taking the sample points to be midpoints. What does the riemann sum represent? Illustrate with a diagram.

I get $.5(f(.5)+f(1)f(1.5)f(2))$ which is the incorrect answer. What am I doing wrong? To find $\Delta x$ I subtract the beginning from the end which is 2 and divide by the number of intervals which is 4 which gives .5 which is what I used but gave me an incorrect answer.

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Draw a shortish line. Label one end $0$, the other end $2$. We want to divide the interval into $4$ parts, so the inner division points are $0.5$, $1$, and $1.5$. Now look at the $4$ intervals. Their midpoints are $0.25$, $0.75$, $1.25$, $1.75$. That's where you will evaluate your function. So the answer should be $(0.5)[f(0.25)+f(0.75)+f(1.25)+f(1.75)]$. – André Nicolas Nov 1 '11 at 0:48
It never hurts to double check your title for obvious typos. And capitalizing Riemann's name is usually better... – Mariano Suárez-Alvarez Nov 1 '11 at 0:50
I did a little editing for spelling, punctuation, formatting, etc. Please check that I didn't mess up the meaning. – Gerry Myerson Nov 1 '11 at 1:25

Your $\Delta x$ is correct, but your midpoints are not.
Your intervals are $[0,0.5],[0.5,1],[1,1.5],[1.5,2]$.
That means the midpoints will be $0.25,0.75,1.25,$ and $1.75$.
That's not what you wrote in your question: "$(f(.5)+f(1)f(1.5)f(2))$" – AMPerrine Nov 1 '11 at 0:51