Respuesta :
Answer:
Calculators from the beginning of the third week to the end of the fourth week = 4048.
Step-by-step explanation:
We know that the rate of production of these calculators after t weeks is given by
[tex]\frac{dx}{dt} =5000(1-\frac{100}{(t+10)^{2}})[/tex]
To find the number of calculators that have been produced in a period, we need to take the integral of the function above; the desired time is t=2 (beginning of third week) to t=4 (end of the fourth week). Therefore, the number of calculators produced in the given time is
[tex]\int\limits^4_2 {\frac{dx}{dt} } \, dt = \int\limits^4_2 {5000(1-\frac{100}{(t+10)^{2} }) } \, dt [/tex]
Substitute [tex]t+10=u[/tex] and [tex]dt=du[/tex], observe that the limits of integration will change
[tex]\int\limits^4_2 {\frac{dx}{dt} } \, dt => \int\limits^{14}_{12} {\frac{du}{dt} } \, dt[/tex]
[tex]5000\int\limits^{14}_{12} { 1-\frac{100}{u^{2} } } \, du[/tex]
[tex]5000(u+100u^{-1})\left \{ {{14} \atop {12}}\right.\\5000(2+\frac{100}{14}-\frac{100}{12} )\\4047.62[/tex] ≈ 4048
The number of calculators produced from the beginning of the third week to the end of the fourth week is; 4048
How to solve first order differential equations?
We are given the rate of production of the calculators after t weeks as;
dx/dt = 5000[1 − (100/(t + 10)²)] calculators/week
To find the number of calculators that have been produced in a period, we will integrate the above function between t = 2 to t=4.
Thus, the number of calculators produced in the given time is;
[tex]\int\limits^4_2 {\frac{dx}{dt}} \, dt = \int\limits^4_2 {5000(1 - \frac{100}{(t + 10)^{2} } } \, dt[/tex]
If we put t + 10 for u and du for dt, it means that our limits of integration will be; u = 14 and 12. Thus;
[tex]5000\int\limits^{14}_{12} {1 - \frac{100}{u^{2} } } \, du[/tex]
Integrating this with a calculator gives 4048 calculators
Read more on differential equations at; https://brainly.com/question/18760518
