Oszacuj dokładność wzoru przybliżonego
| x2 | x3 | |||
ln(1−x) ≈ −x − | − | , dla |x| ≤ 0.1 | ||
| 2 | 3 |
| −6 | ||
f(IV)(x) = −2(−3)(1−x)−4(−1) = −6(1−x)−4, f(IV)(c) = −6(1−c)−4 = | ||
| (1−c)4 |
| x2 | 2x3 | |||
ln(1−x) = −x − | − | + R4(x), | ||
| 2 | 3! |
| −6x4 | ||
gdzie R4 = | ||
| (1−c)4*4! |
| x2 | 2x3 | |||
ln(1−x) = −x − | − | + R4(x), | ||
| 2 | 3! |
| x2 | 2x3 | |||
| ln(1−x) + x + | + | | = |R4(x)| | ||
| 2 | 3! |
| −6x4 | 6x4 | x4 | ||||
|R4(x)| = | | | = | | | = | | | | |||
| (1−c)4*4! | (1−c)4*4! | (1−c)4*4 |
| 1 | 1 | 1 | ||||
|x| ≤ | , czyli: c ∊(− | ;0) lub (0; | ) | |||
| 10 | 10 | 10 |
| 1 |
|
| ||||||||||||||||||||||
R4( | ) = | | | ≤ | | |, czyli: | |||||||||||||||||||||
| 10 | (1−c)4*4 |
|
| 1 | 1 | 1 | 1 | ||||||||||||||
R4( | ) ≤ | = | = | = | |||||||||||||
| 10 | 4*104 * (9/10)4 |
| 4*94 |
| 1 | |
<< oszacowanie, dobrze? | |
| 26244 |
i jeszcze mam takie coś:
policzyć pole ograniczone krzywymi:
x = 1,
| 5 | ||
y= | π | |
| 6 |
| 5π | ||
z góry ograniczone przez | , z dołu przez arcctg(x), z lewej: | |
| 6 |
| 5π | ||
arcctg(x) | ||
| 6 |
| 5π | ||
ctg( | ) = x | |
| 6 |
| π | ||
ctg(U{π− | ) = x | |
| 6 |
| π | ||
−ctg( | ) = x | |
| 6 |
| 5π | ||
z góry: y = | ||
| 6 |
| 5π | ||
∫−√31 ( | − arcctg(x)) dx = | |
| 6 |
| 5π | 5πx | |||
∫ | dx = | |||
| 6 | 6 |
| u = arcctgx, v' = 1 | x | ||||||||||||
∫arcctgx dx= | = xarcctgx + ∫ | dx = | |||||||||||
| 1+x2 |
| 1 | 2x | 1 | ||||
xarcctgx + | ∫ | dx = xarcctgx + | ln(1+x2) + C | |||
| 2 | 1+x2 | 2 |
| 5π | ||
∫−√31 ( | − arcctg(x)) dx = | |
| 6 |
| 5πx | 1 | ||
− xarcctgx − | ln(1+x2) ↕−√31 = | ||
| 6 | 2 |
| 1*5π | 1 | |||
= | − 1*arcctg(1) − | ln(1+12) − | ||
| 6 | 2 |
| √3*5π | 1 | ||
− (−√3)*arcctg(−√3) − | ln(1+(−√3)2) = | ||
| 6 | 2 |
| 5π | π | 1 | |||
− | − | ln(2) − | |||
| 6 | 4 | 2 |
| √3*5π | 5π | 1 | |||
+√3 * | − | ln(4) = | |||
| 6 | 6 | 2 |
| 5π | π | 1 | 5π | π | 1 | ||||||
− | − | (ln(2*4) = | − | − | (ln(8) | ||||||
| 6 | 4 | 2 | 6 | 4 | 2 |
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