$$\begin{aligned} & \int x^3 \sqrt{3-x^2} d x \\ & 3-x^2=t^2 \\ & -2 x d x=2 t d t \\ & =-\int t^2\left(3-t^2\right) d t=\int t^4-3 t^2 d t \\ & =\frac{t^5}{5}-t^3+c \end{aligned}$$

$$\begin{aligned} & f(x)=\frac{\left(3-x^2\right)^{\frac{5}{2}}}{5}-\left(3-x^2\right)^{\frac{3}{2}}+c \\ & \because f(\sqrt{2})=-\frac{4}{5} \\ & \Rightarrow \quad-\frac{4}{5}=\frac{1}{5}-1+c \Rightarrow c=0 \\ & \begin{aligned} \therefore \quad f(x) & =\frac{\left(3-x^2\right)^{\frac{5}{2}}}{5}-\left(3-x^2\right)^{\frac{3}{2}} \\ \text { Now } f(1) & =\frac{2^{\frac{5}{2}}}{5}-2^{\frac{3}{2}}=2^{\frac{3}{2}}\left[\frac{2}{5}-1\right] \\ & =2 \sqrt{2}\left(-\frac{3}{5}\right)=-\frac{6}{5} \sqrt{2} \end{aligned} \end{aligned}$$