The average value of an integrable function on an interval [a,b] is defined as:

`"Avg"= 1/(b-a) int_(a)^(b) f(x)dx `

Find the average value of `f(t)= t sin(t^2) dt ` on [0,10]:

`"Avg"= 1/10 int_0^10 t sin(t^2) dt `

Let `u=t^2, du=2tdt ==> tdt=du/2 ` ; also x=0 ==> u=0 and x=10 ==> u=100 so we can rewrite as:

` =1/10 int_0^100 sin(u) 1/2 du ` or factoring out the constant:

`=1/20 int_0^100 sin(u)du `

`=1/20[ -cos(u) |_0^100 ] `

`=1/20[-cos(100)+cos(0)] `

`~~.0069 `

---------------------------------------------------------------------------

The average value of f(t) on [0,10] is approximately .0069

--------------------------------------------------------------------------