posted by Justin .
Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places.
int_2^10 2 sqrt(x^2+5)dx text(, ) n=4
Explain what you mean by "n" and "text(, )"
Probably no one has responded to your questions because your notation is unfamiliar.
In general midpoint rule is
Xo to Xn is approximated by rectangles of base delta x = Xk - Xk-1 and height f([Xk+Xk-1]/2)
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