The function is tsin(t^2) on the interval [0,10]
So it'd end up being 1/10 the integral from 0 to 10 of tsin(t^2), but that's where I'm stuck. How in the world would I take the antiderivative of that function?
Are you sure you arent finding the average value of sin(t^2) on the interval [0,10] ?
Then the average value is:
$\displaystyle \int_{0 }^{10 } tSin( { t}^{2 }) $
then substitute u = t^2 and solve.
You should get around .0688.
If you are finding the average value of tsin(t^2), the average value is:
$\displaystyle \int_{0 }^{10 } { t}^{ 2} Sin( { t}^{2 }) $
and this integral is now very difficult and involves the Fresnel integral. The value here is about -4.01.