I am studying multivariable calculus and I'm currently doing operations of vector valued functions of the form:
(t) = f(t)i + g(t)j +h(t)k
I came upon a theorem in my textbook that says the definite integral of the function is the definite integral of the components of , which makes perfect sense to me. However, what doesn't make sense is I've also found Calculus resources online that say the Fundamental Theorem of Calculus can be extended to vector valued functions like . Is it because the definite integral of is just the sum of the integrals of its components, which are itself real-valued functions (multiplied by unit vectors) and not vector valued functions? Or is there a way to extend the Mean Value Theorem to vector valued functions to make the Fundamental Theorem proof easily doable? Any help would be greatly appreciated. Thank you!
Consider the vector-valued function $\mathbf r : \mathbb R \to \mathbb R^3$ defined by $\mathbf r(t) = \langle f(t), g(t), h(t) \rangle$ for some integrable functions $f, g,$ and $h.$ By definition, the definite integral of $\mathbf r(t)$ over $[a, b]$ is $$\int_a^b \mathbf r(t) \, dt = \biggl \langle \int_a^b f(t) \, dt, \int_a^b g(t) \, dt, \int_a^b h(t) \, dt \biggr \rangle.$$ By the Fundamental Theorem of Calculus in one variable, we have that $$\biggl \langle \int_a^b f(t) \, dt, \int_a^b g(t) \, dt, \int_a^b h(t) \, dt \biggr \rangle = \langle F(b) - F(a), G(b) - G(a), H(b) - H(a) \rangle \phantom{Give me some space!}$$ $$\phantom{All right! Cool it. You can have it!!} = \langle F(b), G(b), H(b) \rangle - \langle F(a), G(a), H(a) \rangle = \mathbf R(b) - \mathbf R(a),$$ where $\mathbf R(t)$ is an antiderivative of $\mathbf r(t)$ determined by the antiderivatives $F,$ $G,$ and $H$ of $f,$ $g,$ and $h,$ respectively. Likewise, by definition, the derivative of the vector-valued function $\int_a^t \mathbf r(u) \, du$ is $$\frac{d}{dt} \int_a^t \mathbf r(u) \, du = \biggl \langle \frac d {dt} \int_a^t f(u) \, du, \frac d {dt} \int_a^t g(u) \, du, \frac d {dt} \int_a^t h(u) \, du \biggr \rangle = \langle f(t), g(t), h(t) \rangle = \mathbf r(t).$$ Consequently, both parts of the Fundamental Theorem of Calculus hold for vector-valued functions by virtue of the Fundamental Theorem of Calculus for functions of one variable and the fact that limits, derivatives, and integrals for vector-valued functions are taken componentwise.