# The rate of expenditure for maintenance of a particular machine is given by M'(x)=(sqrt (x^2+8x))*(2x+8), where x is time measured in years.Aggregate maintenance costs after the 4th year $540.a)Find aggregate maintenance function. b) How many years must pass before the aggregate maintenance costs reach$2000?

The rate of expenditure in year x is, M'(x)

M'(x) = (sqrt(x^2+8x))*(2x+8)

If we integrate this we get the aggregate maintenance function over the years, let's name it as M(x)

M(x) = int(sqrt(x^2+8x))*(2x+8) dx

this can be easily doen through a simpel subsitution,

let's say,

u = x^2+8 then, du = (2x+8)dx

Therefore the integral changes

M(x) = intsqrtu*du

M(x) = u^(3/2)/(3/2)+C

C is the constant, we have to find it through given data.

M(x) = (2(x^2+8)^(3/2))/3+C

now we are given one data that, after 4 years, the aggregate maintenance cost is 540.

Therefore,

540 = (2(4^2+8)^(3/2))/3+C

540 = 78.3837 + C

C = 461.6163

a)therefore the aggregate total function is,

M(x) = (2(x^2+8)^(3/2))/3+461.6163

b) To find the year in which aggregate cost are equal to 2000, you have to solve for x

2000 = (2(x^2+8)^(3/2))/3+461.6163

(2(x^2+8)^(3/2))/3 = 1538.3837

(x^2+8)^(3/2) = 2307.57555

x^2+8 = 174.624

x = 12.908

Therefore the 12.9 (or 13 years) should pass for aggregate costs to reach 2000.