let the annual revenue be A(x)
then
A ' (x) = R(x) = 8.2x^2 + 23.1x
A(x) = (1/3)(8.2)x^3 + (23.1/2)x^2 + C
= (41/15) x^3 + (23/20)x^2 + c
at year 2000, t = 0 and c = 530
at year 2002, t = 2
A(2) = (41/15)(8) + (23/20)(4) + 530
A(5) = (41/15)(125) + (23/20)(25) + 530
A(5) - A(2)
= [ (41/15)(125) + (23/20)(25) + 530 ] - [(41/15)(8) + (23/20)(4) + 530]
= $ 343.95
Part I
The change in annual revenue in thousands of dollars for the Ozark Zip Line Corporation has been modeled by the function.
R(x) = 8.2x^2 + 23.1x
where x is the number of years after 2000. To find the total revenue gained between the years 2002 and 2005, start by choosing a constant (C)530
Part II
Build the definite integral that is used to find the total revenue gained between the years 2002 and 2005
Part III
Apply the power rule to integrate R(x) over the given interval, and round your answer to the nearest whole tens of dollars.
1 answer