Since YouTube first became available to the public in mid-2005, the rate at which video has been uploaded to the site can be approximated by

v(t) = 525,600(0.42t2 + 2.7t − 1) hours of video per year (0.5 ≤ t ≤ 7),
where t is time in years since the start of 2005.

(a) Find an expression for the total number of hours
V(t) of video at time t (starting from zero hours of video at t = 0.5).

V(t) =

(b) Use the answer to part (a) to estimate the total number of hours of video uploaded by the start of 2010.

Question

Since YouTube first became available to the public in mid-2005, the rate at which video has been uploaded to the site can be approximated by

v(t) = 525,600(0.42t2 + 2.7t − 1)  hours of video per year (0.5 ≤ t ≤ 7),
where t is time in years since the start of 2005.

(a) Find an expression for the total number of hours 
V(t) of video at time t (starting from zero hours of video at t = 0.5).

V(t) =  
 
(b) Use the answer to part (a) to estimate the total number of hours of video uploaded by the start of 2010.

anonymous · Answer

For part a, isn't the answer the given function?

geerky42 · Answer

@ArkGoLucky None, we are given the rate of hours of video uploaded. What we want is total number of hours from beginning.

geerky42 · Answer

So since $v(t)=V'(t)$ and we want to find $V(t)$, we would need to take integral of $v(t)$, from initial time $t_o=0.5$ to desired time $t$.

Which would looks like $\displaystyle\int_{0.5}^tv(x)~dx \~\\displaystyle= \int_{0.5}^t 525,600(0.42x^2 + 2.7x − 1)~dx\~\=\cdots~?$

geerky42 · Answer

@rmwendt11