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. f(t) =

(b) If the trend continued, what was the Internet advertising revenue in 2007?

Answer 1

Answer:

a) $\bf f(t)=\displaystyle\frac{0.82t^2}{2}+1.14t+5.9$

b) 21.85 billion $

Step-by-step explanation:

Let f(t) be the advertising revenue in year t.

(a) Find an expression f(t) giving the advertising revenue in year t.

Since R(t) is the rate of growth of f(t), then R(t) = f'(t) and

$\bf f(t)=\int_{0}^{t}R(x)dx=\int_{0}^{t}(0.82x+1.14)dx=\\\\=0.82\int_{0}^{t}xdx+1.14\int_{0}^{t}dx=\displaystyle\frac{0.82t^2}{2}+1.14t+C$

where 0 ≤ t ≤ 4 and C is a constant.

The advertising revenue in 2002 (t=0) was $5.9 billion, so f(0) = 5.9. Since f(0) = C, then C=5.9 and  

$\bf \boxed{f(t)=\displaystyle\frac{0.82t^2}{2}+1.14t+5.9}$

(b) If the trend continued, what was the Internet advertising revenue in 2007?

if t=0 represents the year 2002 then t=5 represents the year 2007 and the revenue was f(5)

$\bf f(5)=\displaystyle\frac{0.82(5^2)}{2}+1.14*5+5.9=\boxed{21.85\;billion}$