# Thread: Revenue Function - Integrals

1. ## Revenue Function - Integrals

The marginal-revenue function for a manufacturer's product is of the form
$\displaystyle \frac{dr}{dq}=\frac{a}{e^q+b}$
for constands a and b, where r is total revenue received when q units are produced and sold.

Find the demand function, and express it in the form p = f(q)

I'm not sure where to even begin, any help is appreciated....thank you!

2. To find the demand function $\displaystyle p$, you must find $\displaystyle r$, then use $\displaystyle p=\frac{r}{q}$.

To solve the integral

$\displaystyle r(q)=a\int\frac{dq}{e^q+b}$,

you use the substitution $\displaystyle u=e^q+b$. Then $\displaystyle du=e^q\,dq=(u-b)\,dq$,so

$\displaystyle r(q)=a\int\frac{dq}{e^q+b}=a\int\frac{du}{u(u-b)}=\frac{a}{b}\int\left(\frac{1}{u-b}-\frac{1}{u}\right)\, du$ ...

3. Evaluate $\displaystyle \int \frac{a}{e^q + b} dq$ to find revenue in terms of quantity sold. You know that if you sell nothing then you make no revenue, so use the fact that q = 0 implies r = 0 to solve for the constant of integration. Since $\displaystyle r = pq = f(q)q$, once you have r you can find f(q).

4. $\displaystyle \int \frac{a}{e^q + b} \, dq$

$\displaystyle a \int \frac{1}{e^q + b} \cdot \frac{e^{-q}}{e^{-q}} \, dq$

$\displaystyle a \int \frac{e^{-q}}{1 + be^{-q}} \, dq$

$\displaystyle u = 1 + be^{-q}$

$\displaystyle du = -be^{-q} \, dq$

$\displaystyle -\frac{a}{b} \int \frac{-be^{-q}}{1 + be^{-q}} \, dq$

$\displaystyle -\frac{a}{b} \int \frac{du}{u}$

$\displaystyle -\frac{a}{b} \ln|u| + C$

$\displaystyle -\frac{a}{b} \ln(1+be^{-q}) + C$