Question

The rate of change of revenue (in dollars per calculator) from the sale of x calculators is R′(x)=(x+1)ln(x+1)R ′ (x)=(x+1)ln(x+1). Find the total revenue from the sale of the first 12 calculators. (Hint: In this exercise, it simplifies matters to write an antiderivative of x+1 as (x+8)2/2(x+8) 2 /2 rather than x2/2+x.)x 2 /2+x.)

Answer 1

Answer:

Total revenue = $\frac{169}{2}\ln(13)-\frac{169}{4}$  dollars

Step-by-step explanation:

$R'(x)=(x+1)\ln(x+1)$

Integrate with respect to $x$.

$R(x)=$ ∫$(x+1)\ln(x+1)$ dx

Take $x+1=u$

Differentiate with respect to $x$

$dx=du$

So,

$R(x)=$ ∫$u\ln(u)\,du$

Use integration by parts: ∫f g dx = f∫ g dx-∫(∫g dx)f' dx

Therefore,

$R(x)=$  $\frac{u^2}{2}\ln(u)$ $-$  ∫$\frac{u^2}{2}\frac{1}{u}\,du$

=  $\frac{u^2}{2}\ln(u)$ $-$  ∫$\frac{u}{2}\,du$

Put $u=x+1$

$R(x)= \frac{(x+1)^2}{2}\ln(x+1)-\frac{(x+1)^2}{4}$

To find total revenue from the sale of the first 12 calculators, put $x=12$

$R(x)= \frac{(12+1)^2}{2}\ln(12+1)-\frac{(12+1)^2}{4}\\\\= \frac{(13)^2}{2}\ln(13)-\frac{(13)^2}{4}\\\\=\frac{169}{2}\ln(13)-\frac{169}{4}$

Total revenue = $\frac{169}{2}\ln(13)-\frac{169}{4}$  dollars