Showing Validity of FTC for Vector-valued Functions

[onie mti]

how can i show that the FTC are valid for vector valued functions, as in

if a function x is a solution of the IVP
x^'=f(x) with x(s)=b
if and only if

x(t) = b + (integral from s to t) [f(x (θ)) d(θ)] for each t

 

[jvicens]

in [s,∞)To show that the fundamental theorem of calculus (FTC) is valid for vector valued functions, we need to understand the properties of vector functions and how they behave under integration.

Firstly, a vector function is a function that maps a set of inputs to a set of vectors. In the context of the given problem, the function x is a vector function that maps a set of inputs to a set of vectors in the form of x(t) = [x1(t), x2(t), ..., xn(t)]. Therefore, the integral of a vector function would result in a vector-valued output.

Now, let's consider the FTC for single-variable functions, which states that if a function f(x) is continuous on the interval [a, b], then the integral of f(x) from a to b is equal to the difference of the antiderivative of f(x) evaluated at b and a, i.e. ∫a^b f(x) dx = F(b) - F(a), where F(x) is the antiderivative of f(x).

We can extend this concept to vector functions by considering each component of the vector separately. In other words, if we have a vector function x(t) = [x1(t), x2(t), ..., xn(t)], then the FTC would hold for each component, i.e. ∫a^b x1(t) dt = X1(b) - X1(a), ∫a^b x2(t) dt = X2(b) - X2(a), ..., ∫a^b xn(t) dt = Xn(b) - Xn(a). Here, X1, X2, ..., Xn are the antiderivatives of x1, x2, ..., xn respectively.

Now, let's consider the given IVP x'=f(x) with x(s)=b. We can rewrite this as a system of equations: x1'=f1(x), x2'=f2(x), ..., xn'=fn(x), where f1, f2, ..., fn are the components of the vector function f(x). By solving this system of equations, we can find the individual components of x(t), which we can then integrate using the FTC. This would result in x(t) = [X1(t), X2(t), ..., Xn(t)] = b + ∫s^t f(x(θ)) dθ,