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Question: https://gyazo.com/acce090bee3f6ccc8346b39e2fe5ab4b

My solution and to the part where I do not understand:

https://gyazo.com/72ab1498de0455f79a0497a2734745fe

I feel I am just computing trapezoid sum rather than midpoint sum. How would I get the midpoint sum?

Also the definite integral would be limit of n approaches infinity, but I do not know if I have to put that since we are given the values.

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    Trapezoid sum is more appropriate for this problem. For your second question about the definite integral, the infinity is not use in numerical method but is an analytical method. – 2017-01-17
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    I noticed that, but how would I find the midpoint sum? I think I could partition it but I am not too sure. – 2017-01-17
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    You don't know the midpoint value and you cannot do midpoint sum. This is a fact and there is no trick. – 2017-01-17
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    Are you saying this question has no solution? – 2017-01-17
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    I think using midpoint sum is not possible. – 2017-01-17
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    You certainly can use the midpoint sum. Just use the second point on the list and every other from there on out. – 2017-01-17

1 Answers 1

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To use the midpoint rule you need the partition $[1,1.8,2.6,3.4,4.2]$ which has four subintervals with midpoints $1.4,2.2,3.0,3.8.$ The length of each subinterval is $0.8,$ so the midpoint approx is $(0.8)[8.6+7.6+6.3+6.5]=23.2.$

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    How do you know to use 0.8 as the $\Delta x$ ? And why four subintervals? My textbook doesn't say much about midpoint sum so I'm kind of confused. – 2017-01-17
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    That is the most intervals you can get using midpoint and the table of values. You could, for example, just use the value at 2.6 (the midpoint of the whole interval), but you would get a worse approximation. – 2017-01-17
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    Ohhhhh, thx I understand now and your solution makes sense. – 2017-01-17
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    @Tinler The data only had values enough to make four subintervals. (use every other $x$ as a midpoint.) – 2017-01-17