| 1 | ||
t = 4 − 2x −−−> 2x = 4−t −−−> x = | (4−t) | |
| 2 |
| 1 | ||
dt = −2dx −−−> dx = − | dt | |
| 2 |
| 1 | 1 | 1 | 1 | |||||
∫ t2 * | (4−t) (− | )dt = | ∫ t2(t−4) dt = | ∫ (t3 − 4t2) dt | ||||
| 2 | 2 | 4 | 4 |
| 1 | ||
− | dt=dx | |
| 2 |
| 1 | ||
− | ∫t2dt | |
| 2 |
| 1 | ||
=− | t3+C | |
| 6 |
| 1 | ||
=− | (4−2x)3+C | |
| 6 |
| 4 | ||
∫ (4 − 2x)2 dx = 4 ∫ (x − 2)2 dx | x − 2 = u ⇒ dx = du | = 4 ∫ u2 du = | u3 + C = | |
| 3 |
| 4 | |
(x − 2)3 + C | |
| 3 |
| 1 | 1 | |||
− | x(4−2x)3+ | ∫(4−2x)3dx | ||
| 6 | 6 |
| 1 | 1 | |||
=− | x(4−2x)3− | (4−2x)4+C | ||
| 6 | 48 |
| 1 | 4 | |||
∫( t3 − 4t2)dt = | t4 − | t3 + C | ||
| 4 | 3 |
| 16x3 | ||
x4− | +8x2+C | |
| 3 |
z wyliczając wyrażenie końcowe Mariusza wyszło mi x4−163x3+4x2−163
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
| Kliknij po więcej przykładów | |
|---|---|
| Twój nick | |