Hello, If possible, may someone take a look at the problem below (note my answers). I am not sure why but I only get the first part right (my answer is seven (7). Thanks. Regards, Chris K.
Follow Math Help Forum on Facebook and Google+
The terms of a Riemann Sum are f(xk)deltax deltax = 2/n x0 = 5 x1 = 5 +2/n xk = 5 +2k/n f(xk) = sqrt(8 + 2k/n) so f(x) = sqrt(3 + 5+2k/n) f(x) = sqrt( 3 +x)
Originally Posted by Calculus26 The terms of a Riemann Sum are f(xk)deltax deltax = 2/n x0 = 5 x1 = 5 +2/n xk = 5 +2k/n f(xk) = sqrt(8 + 2k/n) so f(x) = sqrt(3 + 5+2k/n) f(x) = sqrt( 3 +x) Thanks!
so do you also see why g(x) = sqrt(x) with c = 10 ?
Originally Posted by Calculus26 so do you also see why g(x) = sqrt(x) with c = 10 ? I see that g(x) would be the new limits should g(x) be the anti-derivate of f(x). Am I correct?
No g(x) is simply a horizontal shift of f(x) to the right 3 units If you compute the integrals you get the same results as the area doesn't change. See attachment
View Tag Cloud