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
Calculus - drwls, Saturday, January 19, 2008 at 10:34am
Explain 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 2:30pm
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