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October 31, 2014

October 31, 2014

Posted by **Justin** on Friday, January 18, 2008 at 9:29pm.

int_2^10 2 sqrt(x^2+5)dx text(, ) n=4

- Calculus -
**drwls**, Saturday, January 19, 2008 at 10:34amExplain what you mean by "n" and "text(, )"

Probably no one has responded to your questions because your notation is unfamiliar.

- Calculus -
**Damon**, Saturday, January 19, 2008 at 2:30pmIn general midpoint rule is

integral from

Xo to Xn is approximated by rectangles of base delta x = Xk - Xk-1 and height f([Xk+Xk-1]/2)

so

integral is approximately

sum from k = 1 to k = n of f([Xk+Xk-1]/2) delta x

NOW HERE - You have not told us over what interval

nor do I really understand what f(x) is

like what does 2^10 2 mean?

2^10 is 2 to the tenth. Is this times another two so it is 2^11. Please be careful with parentheses.

Anyway here is how to do the problem.

Divide whatever your interval is by four if n is four then delta x = interval/4

label those five x points Xo X1 X2 X3 X4

calculate f(x) at (X1-Xo)/2 and (X2-X1)/2 and (X3-X2)/2 and (X4-X3)/2

then add them up and multiply by delta x

done

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