Test marketing for a new health-food snack product in a selected area suggests that sales (in thousands of dollars) will increase at a rate given by S'(t) = 40 - 40e^(-0.16t), t months after an aggressive national advertising campaign is begun. Find total sales during the second 12 months of the campaign. (Round to the nearest thousand dollars.)
Second 12 months mean 24 months?
What is the antiderivative of S'(t) = 40 - 40e^(-0.16t), is it 40t - 40e^(-0.16t) ?
Second 12 months mean 24 months?
What is the antiderivative of S'(t) = 40 - 40e^(-0.16t), is it 40t - 40e^(-0.16t) ?
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the antiderivative of S'(t) = 40 - 40e^(-0.16t) is
S(t) = 40t + 250e^(-0.16t) + C
and it appears you are to find the definite integral
of S(t) over the interval from (t = 12) to (t = 24).
S(t) = 40t + 250e^(-0.16t) + C
and it appears you are to find the definite integral
of S(t) over the interval from (t = 12) to (t = 24).