Annual sales of fountain pens in Littleville are 2,000 per year and are increasing by 10% per year. How many fountain pens will be sold over the next five years? Use an integral to solve this problem.
I just need help setting the problem up - thank you!
1.1 times as many pens are sold each year, so after t years the sales will be
2000 * 1.1^t
To get the sum of all sales, do the integral
∫[1,5] 2000 * 1.1^t dt
You need to check on how to integrate
a^t for any constant a
To solve this problem using an integral, we can set up a definite integral to find the total number of fountain pens sold over the next five years.
Let's first define the notation:
Let f(t) be the sales rate of fountain pens at time t (in years).
Given that the sales rate is increasing by 10% per year, we can express f(t) as:
f(t) = 2000 * (1 + 0.10)^t
Here, t represents the time in years. At t=0, we have the baseline sales rate of 2000 fountain pens per year.
To find the total number of fountain pens sold over the next five years, we need to integrate f(t) over the interval from 0 to 5.
∫[0,5] f(t) dt represents the definite integral over the interval from 0 to 5 of the sales rate function f(t), which will give us the total number of fountain pens sold during that time period.
Now, we substitute the expression for f(t) into the integral:
∫[0,5] f(t) dt = ∫[0,5] 2000 * (1 + 0.10)^t dt
To evaluate this integral, we can use the power rule of integration:
∫a^x dx = (1/(ln(a))) * a^x + C
where a is a constant and C is the constant of integration.
Applying the power rule to our integral:
∫[0,5] 2000 * (1 + 0.10)^t dt = 2000 * (1/(ln(1.10))) * (1.10^t) |[0,5]
Now, we evaluate the integral:
2000 * (1/(ln(1.10))) * (1.10^5) - 2000 * (1/(ln(1.10))) * (1.10^0)
Simplifying further:
2000 * (1/(ln(1.10))) * (1.10^5 - 1)
Evaluating this expression will give us the total number of fountain pens sold over the next five years.