Respuesta :
Answer:
$16749.6 the company pay for waste removal during the year 2016.
Step-by-step explanation:
We know that the amount of waste a company produces, [tex]W[/tex], is given by
[tex]W=3.75e^{-0.008t}[/tex] [tex]\frac{tons}{week}[/tex]
where
[tex]t[/tex] is in weeks since January 1, 2016.
We want to find the total waste during the year 2016. Therefore, we compute the definite integral
[tex]\int\limits^{52 }_0 {3.75e^{-0.008t}} \, dt[/tex]
We have a function that describes a rate ([tex]\frac{tons}{week}[/tex]). So the definite integral gave us the net change in the amount of waste between January 1, 2016 and December 31, 2016.
The limit of integration 52 is because there are 52 weeks in a year.
[tex]\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\3.75\cdot \int _0^{52}e^{-0.008t}dt\\\\\mathrm{Apply\:u-substitution:}\:u=-0.008t\\\\3.75\cdot \int _0^{-0.416}-\frac{1}{0.008}e^udu[/tex]
[tex]3.75\left(-\left(-\frac{1}{0.008}\cdot \int _{-0.416}^0e^udu\right)\right)\\\\3.75\left(-\left(-125\cdot \int _{-0.416}^0e^udu\right)\right)\\\\\mathrm{Use\:the\:common\:integral}:\quad \int \:e^udu=e^u\\\\3.75\left(-\left(-125\left[e^u\right]^0_{-0.416}\right)\right)\\\\468.75\left[e^u\right]^0_{-0.416}\\\\468.75\cdot \:0.34031\\\\159.52[/tex]
The waste removal for the company costs $105/ton. Therefore, the company paid
[tex]159.52 \cdot 105=16749.6[/tex]
