Розв'яжіть W = ∫ f(t)dt=[0001t^3-001t^2-03t+4]_0^12approx

$$W = \int f ( t ) d t = [ 0001 t ^ { 3 } - 001 t ^ { 2 } - 03 t + 4 ] _ { 0 } ^ { 12 } \approx$$

Знайти t

$$\left\{\begin{matrix}t=-\sqrt{\frac{С}{f}}\text{; }t=\sqrt{\frac{С}{f}}\text{; }t=-\sqrt{\frac{2W-С}{f}}\text{; }t=\sqrt{\frac{2W-С}{f}}\text{, }&С=0\text{ and }W=0\text{ and }f>0\\t=-\sqrt{\frac{С}{f}}\text{; }t=\sqrt{\frac{С}{f}}\text{; }t=-\sqrt{\frac{2W-С}{f}}\text{; }t=\sqrt{\frac{2W-С}{f}}\text{, }&С=0\text{ and }W=0\text{ and }f<0\\t\in \mathrm{R}\text{, }&W=4\text{ and }f=0\text{ and }С=4\end{matrix}\right,$$

Знайти f

$$\left\{\begin{matrix}f=\frac{С}{t^{2}}\text{, }&W=0\text{ and }t\neq 0\\f\in \mathrm{R}\text{, }&W=4\text{ and }t=0\text{ and }С=4\end{matrix}\right,$$