Suppose I have a mass with equation of motion described by:
$x^{''}(t) = F(t) - 1$, $0<t<T$, all initial conditions equal to zero
$F(t)$ is some unknown force
My text claims that the equation of the position of the object is
$x(t) = \int_0^t (t-\tau) F(\tau) d\tau - t^2/2$
Does anyone see how the first term of this solution was derived?
Integrating once (using $x'(0)=0$), from zero to $\tau$, we have $$ x'(\tau)=\int_0^\tau F(s)\,ds-\tau. $$ Integrating once more (using $x(0)=0$), from zero to $t$, we find $$ x(t)=\int_0^t \int_0^\tau F(s)\,ds\,d\tau-t^2/2. $$ In the integral, we integrate by parts, $$ \int_0^t 1 \int_0^\tau F(s)\,ds\,d\tau=\Bigl[(\tau-t)\int_0^\tau F(s)\,ds\Bigr]_0^t-\int_0^t (\tau-t)F(\tau)\,d\tau=\int_0^t(t-\tau)F(\tau)\,d\tau. $$ In the last step, the expression inside brackets are zero.