Step 1 — Zeros of v(t)
\((t-1)(t-3)=0\Rightarrow t=1,3\). Sign: \(+\) on \([0,1)\), \(-\) on \((1,3)\), \(+\) on \((3,4]\).
Step 2 — Split Integral
\[D=\int_0^1(t^2-4t+3)\,dt+\int_1^3(4t-t^2-3)\,dt+\int_3^4(t^2-4t+3)\,dt\]
Step 3 — Evaluate Each Piece
\(I_1=\left[\tfrac{t^3}{3}-2t^2+3t\right]_0^1=\tfrac{1}{3}-2+3=\tfrac{4}{3}\)
\(I_2=\left[-\tfrac{t^3}{3}+2t^2-3t\right]_1^3=(-9+18-9)-(-\tfrac{1}{3}+2-3)=0+\tfrac{4}{3}=\tfrac{8}{3}\)
(taking absolute value of negative part)
\(I_3=\left[\tfrac{t^3}{3}-2t^2+3t\right]_3^4=(\tfrac{64}{3}-32+12)-(9-18+9)=\tfrac{4}{3}-0=\tfrac{4}{3}\)
\(D=\tfrac{4}{3}+\tfrac{8}{3}+\tfrac{4}{3}=\dfrac{16}{3}\)