5.6 Integrals involving exponential and logarithmic functions  (Page 2/4)

To find the price–demand equation, integrate the marginal price–demand function. First find the antiderivative, then look at the particulars. Thus,

Using substitution, let $u=\mathrm{-0.01}x$ and $du=\mathrm{-0.01}dx.$ Then, divide both sides of the du equation by −0.01. This gives

The next step is to solve for C . We know that when the price is $2.35 per tube, the demand is 50 tubes per week. This means

Now, just solve for C :

Thus,

If the supermarket sells 100 tubes of toothpaste per week, the price would be

The supermarket should charge $1.99 per tube if it is selling 100 tubes per week.

Again, substitution is the method to use. Let $u=1-x,$ so $du=\mathrm{-1}dx$ or $\text{−}du=dx.$ Then ${\displaystyle \int e^{1-x}dx=\text{−}{\displaystyle \int e^{u}du}}.$ Next, change the limits of integration. Using the equation $u=1-x,$ we have

The integral then becomes

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We have

Then, at $t=0$ we have $Q(0)=10=\frac{1}{\text{ln}3}+C,$ so $C\approx 9.090$ and we get

At time $t=2,$ we have

After 2 hours, there are 17,282 bacteria in the dish.

Let $G\left(t\right)$ represent the number of flies in the population at time t . Applying the net change theorem, we have

There are 122 flies in the population after 10 days.