598
Chapter 5 | Integration
p (50) =1.5 e −0.01(50) + C =2.35.
Now, just solve for C :
C =2.35−1.5 e −0.5 = 2.35 − 0.91 =1.44.
Thus,
p ( x ) =1.5 e −0.01 x +1.44. If the supermarket sells 100 tubes of toothpaste per week, the price would be p (100) =1.5 e −0.01(100) +1.44=1.5 e −1 + 1.44 ≈ 1.99. The supermarket should charge $1.99 per tube if it is selling 100 tubes per week.
Example 5.41 Evaluating a Definite Integral Involving an Exponential Function
Evaluate the definite integral ∫ 1 2
e 1− x dx .
Solution Again, substitution is the method to use. Let u =1− x , so du =−1 dx or − du = dx . Then ∫ e 1− x dx =− ∫ e u du . Next, change the limits of integration. Using the equation u =1− x , we have u =1−(1) =0 u =1−(2)=−1. The integral then becomes ∫ 1 2 e 1− x dx =− ∫ 0 −1 e u du = ∫ −1 0 e u du = e u | −1 0 = e 0 − ⎛ ⎝ e −1
⎞ ⎠ =− e −1 +1.
See Figure 5.38 .
This OpenStax book is available for free at http://cnx.org/content/col11964/1.12
Made with FlippingBook - professional solution for displaying marketing and sales documents online