# Evalute the integral I in stages as followed: In a large industrial city the amount of pollutant released into the atmosphere varies seasonally.  A model for this process is P(t) = 3+2cos(t+1/26)pi where P(t) represents the number of tonnes of pollutant released per week in the t–th week of the year with t = 0 corresponding to 1st January. Thus I = ∫ [3+2cos(t+1/26)pi]dt          will calculate the total number of tonnes of pollutant released into the atmosphere during January (a)       Evaluate the integral I in stages as follows: 1)    Find J = ∫ 3dt 2)  Find K = ∫ 2cos(t+1/26)pi dt using the substitution u = (t+1/26)pi 3) Find J +K Note: All the integral limits are from 4 to 0

## Expert Answers

1) You need to evaluate the definite integral J such that:

`J = int_0^4 3dt`

You need to bring the constant term 3 in front of integral sign such that:

`J= 3 int_0^4 dt =gt J = 3t|_0^4`

`J = 3(4 - 0) = 12`

2) You need to evaluate the definite integral K such that:

`K = int_0^4 2cos(((t+1)pi)/26) dt`

You should come up with the substitution `((t+ 1)pi)/26 = u.`

Differentiating both sides yields:

`(pi/26)dt = du =gt dt = (26du)/pi`

Use this substitution under integral sign such that:

`K = int_0^4 2cos u *(26du)/pi`

`K= 52/pi*int_0^4 cos u du =gt K = 52/pi*sin u|_(u_1)^(u_2)`

`K = 52/pi* sin(t + 1/26)pi|_0^4`

`K = 52/pi*(sin (5pi/26) - sin (pi/26))`

`K = 52/pi*(0.568 - 0.120)`

`K = 7.415`

3) You need to evaluate J+K such that:

`J+K = 12 + 7.415 = 19.415`

Hence, evaluating the definite integral following the steps 1),2),3) yields `J+K = 19.415.`

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