Całka z pierwiastkiem
Fryta: Jak te całkę policzyć pomocy
10 mar 11:41
ICSP: x + 1 = t
6
dx = 6t
5
| | 1 − √x+1 | | 1 − t3 | |
∫ |
| dx = ∫ |
| * 6t5 dt = ... |
| | 1 + 3√x+1 | | 1 + t2 | |
10 mar 11:45
chichi:
Miałeś sam to zrobić
10 mar 11:50
Fryta: Chichi ale nie umiem XDDDD
10 mar 11:51
Mila:
| (1−t3)*t5 | | t8−t5 | |
| =− |
| |
| (t2+1 | | t2+1 | |
| | (t8−t5) | |
1) −6∫ |
| dt= ... |
| | (t2+1) | |
Wykonujemy dzielenie:
(t
8−t
5) : (t
2+1)= t
6−t
4−t
3+t
2+t−1
−(t
8+t
6)
======
−t
6−t
5
−(−t
6−t
4)
==========
−t
5+t
4
−( −t
5−t
3)
=======
t
4+t
3
− (t
4+t
2)
=======
t
3−t
2
−( t
3+t)
=======
−t
2−t
−(−t
2−1)
======
−t+1 reszta⇔
| (t8−t5) | | −t+1 | |
| = (t6−t4−t3+t2+t−1)+ |
| |
| (t2+1) | | t2+1 | |
| | (t8−t5) | | −t+1 | |
2) −6∫ |
| dt=−6∫( t6−t4−t3+t2+t−1+ |
| ) dt= |
| | (t2+1 | | t2+1 | |
| | −t+1 | |
=−6[∫( t6−t4−t3+t2+t−1) dt+∫ |
| ) dt= |
| | t2+1 | |
teraz licz 2 całki
10 mar 20:33