Question

PLEASE HELP URGENT : The rate at which customers arrive at a counter to be served is modeled by the function F defined by ​F(t)=10+6cos((t)/(\pi )) 0<=t<=60 where​ F(t) is measured in customers per minute and t is measured in minutes. To the nearest whole​ number, how many customers arrive at the counter over the 60​-minute ​period?

Answer 1

606 customers arrive at the counter over the 60 minute period.

Step-by-step explanation:

Given : The rate at which customers arrive at a counter to be served is modeled by the function F defined by ​
`F(t)=10+6\cos((t)/(\pi))`
`0\leq t\leq 60` where​ F(t) is measured in customers per minute and t is measured in minutes.

To find : How many customers arrive at the counter over the 60​-minute ​period?

Solution :

We are going to integrate the function in the interval
`0\leq t\leq 60` with respect to time.

Function ​
`F(t)=10+6\cos((t)/(\pi))`

Integrate w.r.t t in the interval
`0\leq t\leq 60`

`\int\limits^(60)_(0) {10+6\cos((t)/(\pi))} \, dt`

`=[10t+6\pi\sin((t)/(\pi))}]^(60)_(0)`

`=10(60)+6\pi\sin((60)/(\pi))-0`

`=600+6.167`

`=606.167`

Approximately, 606 customers arrive at the counter over the 60 minute period.

Answer 2

Answer: 606 costumers ( Approx )

Step-by-step explanation:

Here, the function that shows the the rate at which customers arrive at a counter,

`F (t)=10+6cos((t)/(\pi))`

For 0 ≤ t ≤ 60 minutes,

The number of costumers,

`\int_(0)^(60) F(T)`

`\int_(0)^(60) 10+6cos((t)/(\pi)) dt` -----(1)

Let
`(t)/(\pi)=u`

⇒
`t=\pi u`

⇒
`dt=\pi du`

By substituting this on equation (1),

`=\pi \int_(0)^(60) 10+6 cos u du`

`=\pi [ 10u + 6 sin u ]_(0)^(60)`

`=\pi [ 10 (t)/(\pi) + 6 sin ((t)/(\pi)) ]_(0)^(60)`

`=\pi [ 10 (60)/(\pi) +6 sin ((60)/(\pi))- 0]`

`=600+6.16735797767`

`=606.16735797767\approx 606`