Answer
True
Work Step by Step
Step 1
The Fundamental Theorem of Calculus for vector-valued functions is similar to what we have seen for single-variable calculus: \[ \int_{a}^{b} \mathbf{r}(t) \, dt = \left[\mathbf{R}(t)\right]_{a}^{b} = \mathbf{R}(b) - \mathbf{R}(a) \] where \(\mathbf{R}(t)\) is the antiderivative of \(\mathbf{r}(t)\).
Step 2
Then if \(\mathbf{b} = t\) and we take the derivative of it: \[ \frac{d}{dt} \left(\int_{a}^{t} \mathbf{r}(u) \, du\right) = \frac{d}{dt} \left(\mathbf{R}(t) - \mathbf{R}(a)\right) = \mathbf{r}(t) - \mathbf{0} = \mathbf{r}(t) \] since \(\mathbf{r}(t)\) is the derivative of \(\mathbf{R}(t),\) and \(\mathbf{R}(a)\) is a constant vector when we plug in \(a\).
Result: True statement
