integration of vector-valued functions

A vector-valued function r(t) outputs not a number but an arrow — for example the position of a particle at time t, written r(t) = (x(t), y(t), z(t)). Integrating it asks: if I know the velocity arrow at every instant, what is the displacement arrow? The pleasant answer is that integrating a vector is no harder than integrating each of its components separately; the integral of r(t) is just the vector whose components are the integrals of x(t), y(t), z(t).

This works because integration is a linear operation, and a vector is just a bundle of numbers carried along the fixed coordinate directions. So the integral from a to b of r(t) dt equals (integral of x dt, integral of y dt, integral of z dt). The vector Fundamental Theorem of Calculus follows directly: the definite integral of velocity v(t) from time a to b gives the net displacement r(b) - r(a), exactly as in one dimension but now as an arrow. An indefinite vector integral carries a constant vector of integration, fixed by an initial condition such as the starting position.

This is the language of kinematics and dynamics: integrate acceleration to get velocity, integrate velocity to get position, each step a componentwise integral. The one genuine subtlety is that componentwise integration is only that simple in a fixed (Cartesian) basis. In polar, cylindrical, or spherical coordinates the basis vectors themselves change with position, so you cannot blindly integrate the components — you must account for the moving frame, which is where the scale factors and the machinery of curvilinear coordinates enter.