Answer
$\text{displacement = -6}$
$\text{distance = $\frac{13}{2}$}$
Work Step by Step
$\text {It is given that}$
\begin{align}
a(t) = -2m/s^2; v_0=3m/s; 1\leq t \leq 4
\end{align}
$\text {First, we have to find v(t):}$
\begin{align}
v(t) = \int a(t) \ dt = -2t + C
\end{align}
$\text {By applying the given initial condition:}$
\begin{align}
v(0) = -2 \times 0 +C = 3 \Rrightarrow C = 3 \Rrightarrow v(t) = -2t+3
\end{align}
$\text {Now, we can calculate the displacement and distance:}$
\begin{align}
& displacement = \int_1^4 (3-2t) \ dt = \left[ 3t-t^2\right]_1^4 = -6 \\
& distance = \int_1^4 |3-2t| \ dt = \int_1^{1.5} (3-2t) \ dt + \int_{1.5}^4 (-3+2t) \ dt =\\
& = \left[ 3t-t^2\right]_1^{1.5} + \left[ -3t+t^2\right]_{1.5}^4 = \frac{1}{4}+\frac{25}{4} = \frac{13}{2}
\end{align}
