Final answer:
To evaluate the integral of the vector function ∑(cos(3t)i sin(4t)j t²k) dt, we integrate each component separately, yielding ((1/3)sin(3t) + c)i - ((1/4)cos(4t) + c)j + ((1/3)t³ + c)k with c representing the constant of integration for each component.
Step-by-step explanation:
The question involves evaluating a vector integral. The integral of a vector function is performed by integrating each component of the vector separately. So, we have three distinct integrals to evaluate:
- ∑ cos(3t) dt for the i component
- ∑ sin(4t) dt for the j component
- ∑ t² dt for the k component
The indefinite integrals of these functions are as follows:
- For the i component: ∑ cos(3t) dt = (1/3)sin(3t) + c
- For the j component: ∑ sin(4t) dt = -(1/4)cos(4t) + c
- For the k component: ∑ t² dt = (1/3)t³ + c
Therefore, the integral of the given vector function is:
((1/3)sin(3t) + c)i - ((1/4)cos(4t) + c)j + ((1/3)t³ + c)k.
Each component has its own constant of integration, which we represent by the same letter c for simplicity, but they can differ in value.