Wiki-Quellcode von Lösung Integrale berechnen

Zuletzt geändert von deborakemm am 2026/05/12 16:51

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}}