Finding work via Line Integrals The position of an object with mass $m$ at time is $r(t) = at^2 \vec{i} + bt^3 \vec{j}$, where $0 \leq t \leq 1$.
Part a asks for the force, which I found to be $2ma \vec{i} + 6mbt \vec{j}$, which is correct.
Part b asks for the work done by the force in the given time interval. 
I am not sure how to approach this. I tried integrating the equation from part a from 0 to 1 , but I don't think that is the correct way. Any hints in the right direction are appreciated, thank you! (The correct answer for b is $2ma^2 + 4.5mb^2$)
 A: To compute work what you have to do is
$$\int\limits_{C} \vec{F} \cdot \text{d} \vec{r},$$
which is to say you compute the integral in the given interval of the dot product of the force and the time derivative of the curve. Writing it explicitly
$$W = \int_0^1 (2ma)(2at) + (6mbt)(3bt^2) \, dt.$$
When you do it you will find the desired result.
A: Another approach would be to use the "work-kinetic energy theorem", $ \ W \ = \ \Delta K $
$ = \ K_{final} \ - \ K_{initial} \ . $  The velocity function for the particle is 
$$ \ \vec{v}(t) \ = \ \frac{d}{dt} \ \vec{r}(t) \ = \ \langle \ 2at \ , \ 3bt^2 \ \rangle \  \ ,  $$
so the kinetic energy function is given by
$$ K(t) \ = \  \frac{1}{2}m \ [ \ \vec{v}(t) \centerdot \vec{v}(t) \ ] \ = \ \frac{1}{2}m \ ( \ 4a^2t^2 \ + \ 9b^2t^4 \ ) \ \ . $$
The work done on the particle over the path during $ \ 0 \ \le \ t \ \le \ 1 \ $ is then
$$ W \ = \ K(1) \ - \ K(0) \ = \ \frac{1}{2}m \ \left[ \ ( \ 4a^2 \ + \ 9b^2 \ )  \ - \ 0 \ \right] \ = \ 2ma^2 \ + \ \frac{9}{2}mb^2 \ \ . $$
[At a fundamental level, this really is just another way of expressing the work integral that Fantini presents.]
