Question

U.s. internet advertising revenue grew at the rate of r(t) = 0.82t + 1.14 (0 ≤ t ≤ 4) billion dollars/year between 2002 (t = 0) and 2006 (t = 4). the advertising revenue in 2002 was $5.9 billion.† (a) find an expression f(t) giving the advertising revenue in year t.

Answer 1

Answer:

$f(t) = \int 0.82 t +1.14 dt$

$f(t) = \frac{0.82}{2}t^2 +1.14 t +C$

Where C is a constant, now using the initial condition we got:

$f(2)= 5.9 = 0.41 (2)^2 + 1.14*2 +C$

And solving for C we got:

$C= 5.9 -0.41(4) -1.14*2 = 1.98$

And the function desired for the advertising revenue would be given by:

$f(t) = 0.41t^2 1.14t +1.98$

With f the amount in billions and the the years since 2002 to 2006.

Explanation:

For this case we have the following function who represent the revenue grew rate:

$r(t) = 0.82t +1.14 , 0 \leq t \leq 4$

And we want to calculate the Advertising revenue so we need to integrate the function r(t) and we can use the inidital condition t=0 , f(2)= 5.9 billion.

If we integrate the function we got:

$f(t) = \int 0.82 t +1.14 dt$

$f(t) = \frac{0.82}{2}t^2 +1.14 t +C$

Where C is a constant, now using the initial condition we got:

$f(2)= 5.9 = 0.41 (2)^2 + 1.14*2 +C$

And solving for C we got:

$C= 5.9 -0.41(4) -1.14*2 = 1.98$

And the function desired for the advertising revenue would be given by:

$f(t) = 0.41t^2 1.14t +1.98$

With f the amount in billions and the the years since 2002 to 2006.