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?
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!!