Wiki-Quellcode von Lösung Integrale berechnen
Zuletzt geändert von deborakemm am 2026/05/12 16:51
Verstecke letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
 |
1.1 | 1 | a) |
| 2 | 1. Nullstellen ablesen: {{formula}}x_1 = 0{{/formula}} und {{formula}}x_2 = 2{{/formula}} | ||
| 3 | |||
| 4 | 2. Stammfunktion (mit c=0) bilden: {{formula}}F(x)= \frac{1}{4}x^4-\frac{2}{3}x^3{{/formula}} | ||
| 5 | |||
| 6 | 3. {{formula}}\int_{0}^{2} {x^3-2x^2}\, dx = [F(x)]_{0}^{2} = F(2)-F(0) = \frac{1}{4}2^4-\frac{2}{3}2^3-(\frac{1}{4}0^4-\frac{2}{3}0^3)= -1,33 - 0 = -1,33{{/formula}} | ||
| 7 | |||
| 8 | b) | ||
| 9 | {{formula}}\int_{0}^{3} {2e^{-3x+1}}\, dx = [-\frac{2}{3}e^{-3x+1}]_{0}^{3} = G(3)-G(0) = -\frac{2}{3}e^{-3\cdot 3+1}-(-\frac{2}{3}e^{-3\cdot 0+1}) = 1,81{{/formula}} | ||
| 10 | |||
| 11 | c) | ||
| 12 | {{formula}}\int_{0}^{3} {h(x)}\, dx = [H(x)]_{0}^{3} = H(3)-H(0) = 4,13-3 = 1,13{{/formula}} |