Answer
(a)
$$\dot{x}=7 t / 5$$
$$\int_{3}^{x} d x=\frac{7}{5} \int_{0}^{t} t d t$$
$$x(t)=\frac{7}{10} t^{2}+3$$
(b)
$$\dot{x}=3 e^{-5 t} / 4$$
$$\int_{4}^{x} d x=\frac{3}{4} \int_{0}^{t} e^{-5 t} d t$$
$$x(t)=\frac{3}{20}\left(1-e^{-5 t}\right)+4$$
(c)
$$\ddot{x}=4 t / 7$$
$$\dot{x}(t)-\dot{x}(0)=\frac{4}{7} \int_{0}^{t} t d t$$
$$\dot{x}(t)=\frac{4}{14} t^{2}+5$$
$$\int_{3}^{x} d x=\int_{0}^{t}\left(\frac{4}{14} t^{2}+5\right) d t$$
$$x(t)=\frac{4}{42} t^{3}+5 t+3$$
(d)
$$ \ddot{x}= 8 e^{-4 t} / 3 $$
$$ \dot{x}(t)-\dot{x}(0)=\frac{8}{3} \int_{0}^{t} e^{-4 t} d t $$
$$ \dot{x}(t)=\frac{17}{3}-\frac{8}{12} e^{-4 t} $$
$$ \int_{3}^{x} d x=\int_{0}^{t}\left(\frac{17}{3}-\frac{8}{12} e^{-4 t}\right) d t $$
$$ x(t)=\frac{17}{3} t+\frac{1}{6} e^{-4 t}+\frac{17}{6} $$
Work Step by Step
(a)
$$\dot{x}=7 t / 5$$
$$\int_{3}^{x} d x=\frac{7}{5} \int_{0}^{t} t d t$$
$$x(t)=\frac{7}{10} t^{2}+3$$
(b)
$$\dot{x}=3 e^{-5 t} / 4$$
$$\int_{4}^{x} d x=\frac{3}{4} \int_{0}^{t} e^{-5 t} d t$$
$$x(t)=\frac{3}{20}\left(1-e^{-5 t}\right)+4$$
(c)
$$\ddot{x}=4 t / 7$$
$$\dot{x}(t)-\dot{x}(0)=\frac{4}{7} \int_{0}^{t} t d t$$
$$\dot{x}(t)=\frac{4}{14} t^{2}+5$$
$$\int_{3}^{x} d x=\int_{0}^{t}\left(\frac{4}{14} t^{2}+5\right) d t$$
$$x(t)=\frac{4}{42} t^{3}+5 t+3$$
(d)
$$ \ddot{x}= 8 e^{-4 t} / 3 $$
$$ \dot{x}(t)-\dot{x}(0)=\frac{8}{3} \int_{0}^{t} e^{-4 t} d t $$
$$ \dot{x}(t)=\frac{17}{3}-\frac{8}{12} e^{-4 t} $$
$$ \int_{3}^{x} d x=\int_{0}^{t}\left(\frac{17}{3}-\frac{8}{12} e^{-4 t}\right) d t $$
$$ x(t)=\frac{17}{3} t+\frac{1}{6} e^{-4 t}+\frac{17}{6} $$
