Fundamental theorem of calculus and pullback bundles Let $X$ be a manifold and $\pi:E\rightarrow X$ a vector bundle
over $X$ equipped with a metric $\left\langle \cdot,\cdot\right\rangle $.
Let $f:[0,1]\rightarrow M$ be a smooth map, and consider the pullback
bundle $f^{*}E\rightarrow[0,1]$. This is the vector bundle whose
fiber over $t\in[0,1]$ is
$$
(f^{*}E)_{t}=\left\{ (t,v)\,:\, v\in E_{f(t)}\right\} .
$$
Let $\phi:f^{*}E\rightarrow E$ denote the map $\phi(t,v)=v$.
Let $\nabla$ denote a connection on $E$, and let $D:=f^{*}\nabla$
denote the induced connection on $f^{*}E$. 
$D$ is defined as follows:
suppose $F\in\Gamma(f^{*}E)$ is a section of $f^{*}E$, so that $\phi(F(t))\in E_{f(t)}$
for all $t\in[0,1]$. Fix $s\in[0,1]$, and let $v$ denote any vector
field on $X$ with $v(f(s))=\phi(F(s))$.
Then by definition
$$
(D_{\partial_{t}}F)(s)=\left(s,(\nabla_{\dot{f}(s)}v)(f(s))\right)\mbox{ as elements of }(f^{*}E)_{s},
$$
the points being that the right-hand side of the above expression is independent of the choice of $v$.
Assume now that $f(0)=f(1)$. Then if $F\in\Gamma(f^{*}E)$ then $\phi(F(1))$
and $\phi(F(0))$ both lie in the same vector space $E_{f(0)}$. My question
is: does the following generalization of the fundamental theorem of
calculus always hold?
$$
|\phi(F(1))-\phi(F(0))| \le \int_{0}^{1}|\phi(D_{\partial_{t}}F)(t)|dt.
$$
 A: No. A simple counter example comes when $E$ is not orientable: consider the Möbius strip as a one dimensional bundle over $S^1$. We can give a metric on it and consider the induced Levi-Civita connection. 
If you start with the vector $v$ over $0$, when you parallel transport $v$ around the circle you get $-v$. The norm of the derivative is 0, but the total difference is $2|v|$. 

In the orientable case it is not better. Just take $E$ to be the tangent bundle $TX$ and treat it as a Riemannian manifold. Then parallel transport around a closed loop (for which by definition your RHS expression vanishes) does not necessarily bring a vector back to itself: you will in general pick up a defect related to the area of the loop and the enclosed curvature. 
A: As Willie says, it's not true. It is true if $E$ is a flat trivial vector bundle over $X$. In general however $f$ and $D$ determine a map $\mathrm{Hol}(f,D): E_{f(0)} \rightarrow E_{f(0)}$ called the holonomy map, and if $F$ is parallel then the RHS vanishes but the LHS is $|\mathrm{Hol}(f,D)(F(0))-F(0)|$ which is not necessarily zero.
