Hello All-

I was having some difficulty with the below problem.

∫from 0 to 3 [f(x)] dx = 6

∫from 3 to 5 [f(x)] dx = 4

Find the∫from 0 to 5 [ 3 + 2f(x)] dx

My thought was that the answer would simply be 23. [3+2(4+6)]=23

But this seems too simple. Does anyone know?

Thanks!

Also, one more problem I was having trouble with:

v(t) = (t^2)[ln(t+2)]

Find the acceleration at t=6

All you would have to do I think would be to take the derivative of the velocity function and plug in 6 for t, but I'm not positive about this, and even if this was right, I'm not sure how to take the derivative of this particular function.

Any help is appreciated!!