How do you find the Riemann sum for this integral using right endpoints and n=3 for the integral #int (2x^2+2x+6)dx# with a = 5 and b = 11?

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Answer 1 · 2015-04-12T18:00:23

Right endpoints and n=3 for the integral #int (2x^2+2x+6)dx# with a = 5 and b = 11.

#Delta x = (b-a)/n = (11-5)/3 = 2#

All endpoints:

#a = 5#
#a + Delta x = 5+2 = 7#
#a + 2 Delta x = 7+2 = 9#
#a + 3 Delta x = 9+2 = 11 = b#

So the right endpoints are: #7, 9, 11#
#f(x) = 2x^2+2x+6#

#R_3 = #

#= Delta x [f(7) +f(9) +f(11)] = 2[f(7) +f(9) +f(11)] #

#= 2(2(7)^2+2(7)+6) + 2(2(9)^2+2(9)+6) + 2(2(11)^2+2(11)+6)#