The area under the graph of #f# and above the #x# axis #a <= x <= b#
Can be defined as:
#lim_(nrarroo) sum_(i=1)^n f(x_i"*") Delta x#
Where, for each positive integer #n#, we let #Deltax = (b-a)/n#
And for #i=1,2,3, . . . ,n# we let #x_i"*" # be any number in the #i^"th"# subinterval. That is, #x_i"*" # is in #[a+(i-1)Delta x, a+i Delta x]#. This interval is often denoted #[x_(i-1),x_i]#.
Alternatively
for #i=1,2,3, . . . ,n#, we let #x_i"*" = a+iDeltax#. (These #x_i"*"#'s are the right endpoints of the subintervals.)
In this problem , we have
#f(x) = x^2+sqrt(1+2x)#, and
#a = 7# and #b = 9#, so #Delta x = 2/n#.
Using the first definition above we have
#lim_(nrarroo) sum_(i=1)^n f(x_i"*") Deltax#.
Using the #f# and #Deltax# for this problem we can write
#lim_(nrarroo) sum_(i=1)^n ((x_i"*")^2+sqrt(1+2x_i"*")) 2/n#.
where each #x_i"*" # is in #[7+(2(i-1))/n, 7+(2i)/n]#.
Alternatively
We can use the right endpoints of the subintervals. In which case we get
#lim_(nrarroo) sum_(i=1)^n f(x_i) Delta x#. where for #i=1,2,3, . . . ,n#, we let #x_i = 7+(2i)/n#.
#lim_(nrarroo) sum_(i=1)^n f(7+(2i)/n) Delta x#
Using the #f# and #Deltax# for this problem we can write
#lim_(nrarroo) sum_(i=1)^n [((7+(2i)/n)^2+sqrt(1+2(7+(2i)/n))) 2/n]#
In order to satisfy whoever is evaluating your work, you may need to do some algebra and rewrite this as
#= lim_(nrarroo) sum_(i=1)^n [((49+(28i)/n+(4i^2)/n^2+sqrt(15+(4i)/n))) 2/n]#
#= lim_(nrarroo) sum_(i=1)^n [98/n+(56i)/n^2+(8i^2)/n^3 +2/nsqrt(15+(4i)/n)]#
