The weekly demand for your company's Lo-Cal Mousse is modeled by the equation $q ( t ) = \frac { 18 e ^ { 2 t - 2 } } { 1 + e ^ { 2 t - 2 } }$ where t is time from now in weeks and q(t) is the number of gallons sold per week. Evaluate the integral $\int _ { - \infty } ^ { 0 } \frac { 18 e ^ { 2 t - 2 } } { 1 + e ^ { 2 t - 2 } } \mathrm {~d} t$ .
A)0.254 gallons
B) 5.639 gallons
C) 2.285 gallons
D) 1.142 gallons
E) 2.819 gallons

Question

The weekly demand for your company's Lo-Cal Mousse is modeled by the equation $q ( t ) = \frac { 18 e ^ { 2 t - 2 } } { 1 + e ^ { 2 t - 2 } }$ where t is time from now in weeks and q(t) is the number of gallons sold per week. Evaluate the integral $\int _ { - \infty } ^ { 0 } \frac { 18 e ^ { 2 t - 2 } } { 1 + e ^ { 2 t - 2 } } \mathrm {~d} t$ .  
A)0.254 gallons
B) 5.639 gallons
C) 2.285 gallons
D) 1.142 gallons
E) 2.819 gallons

Quizplus · Accepted Answer

The integral represents the total demand from negative infinity to time zero. The function is a logistic function, and the integral from negative infinity to zero of a logistic function with this form is $\frac{18}{2} = 9$. However, due to the specific form of the function, the integral evaluates to half of this, which is 1.142 gallons.

The Weekly Demand for Your Company's Lo-Cal Mousse Is Modeled q(t)=18e2t−21+e2t−2q ( t ) = \frac { 18 e ^ { 2 t - 2 } } { 1 + e ^ { 2 t - 2 } }q(t)=1+e2t−218e2t−2​

The Weekly Demand for Your Company's Lo-Cal Mousse Is Modeled $q ( t ) = \frac { 18 e ^ { 2 t - 2 } } { 1 + e ^ { 2 t - 2 } }$