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If $f(x) = e^x − 1$, $0 ≤ x ≤ 2$, find the Riemann sum with $n = 4$ correct to six decimal places, taking the sample points to be midpoints.

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closed as off-topic by T. Bongers, TMM, Sujaan Kunalan, nbubis, Jyrki Lahtonen Nov 24 '13 at 6:22

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I would suppose it is intended to be $e^x$ rather than $e\cdot x$. – Karl Kronenfeld Nov 24 '13 at 2:43
yes e to the power of x – user2864644 Nov 24 '13 at 2:47

For the Midpoint Rule, we have:

$I = \dfrac{b-a}{n} \sum_{n=1}^4 f(\overline {x_n}) = \dfrac{1}{2}\left(f(1/4) + f{3/4}+f(5/4)+f(7/4)\right)$

$I = \dfrac{1}{2}\left(-1+e^{1/4}-1+e^{3/4}-1+e^{5/4}-1+e^{7/4}\right) = 4.322985533$

We can see this graphically as:

enter image description here

We can calculate the exact result as:

$$I = \int_0^2 (e^x-1)~dx = -3 + e^2 = 4.389056099$$

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Yea my online homework submission thing says its wrong. – user2864644 Nov 24 '13 at 3:45
Did you only use 6-decimal places as I gave more? – Amzoti Nov 24 '13 at 3:47
yes I did only 6 decimal places. Luckily they give me five tries before I cant answer it anymore. – user2864644 Nov 24 '13 at 4:08
Did it accept it? Check my numbers and see if you get the same. – Amzoti Nov 24 '13 at 4:12
Yea I checked it and its still wrong and out of retries. I will ask my teacher if its right. If so then he will give me the points because the online thing is wrong. – user2864644 Nov 24 '13 at 4:23

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