| 3 | 9 | |||
y= | odpowiedź to jest | π. Mógłby ktoś pomóc? | ||
| √2x2+4x+4 | 2 |
| 3 | ||
limx→∞ | =0 | |
| √2x2+4x+4 |
| 3 | ||
limx→−∞ | =0 | |
| √2x2+4x+4 |
| 9 | ||
V=π−∞∫∞y2dx=π−∞∫∞ | dx | |
| 2x2+4x+4 |
| 9 | 9 | 1 | ||||
π∫ | dx = | π∫ | dx= [(x2+2x+1+1=(x+1)1+1, x+1=t, dx=dt] | |||
| 2x2+4x+4 | 2 | x2+2x+2 |
| 9 | 1 | 9 | 9 | |||||
= | ∫ | dt= | πarctgt= | πarctg(x+1) | ||||
| 2 | t2+1 | 2 | 2 |
| 9 | ||
V= | π[lim{x→∞}arctg(x+1)−limx→−∞arctg(x+1)]= | |
| 2 |
| 9 | π | −π | 9 | π | π | 9π2 | ||||||||
= | π*[ | − | ]= | π*( | + | )= | ||||||||
| 2 | 2 | 2 | 2 | 2 | 2 | 2 |
