Thread: Finding original intergral from given Reimann sum...

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.

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