Lösung Integrale berechnen

a)

1. Nullstellen ablesen: \(x_1 = 0\) und \(x_2 = 2\)

2. Stammfunktion (mit c=0) bilden: \(F(x)= \frac{1}{4}x^4-\frac{2}{3}x^3\)

3. \(\int_{0}^{2} {x^3-2x^2}\, dx = [F(x)]_{0}^{2} = F(2)-F(0) = \frac{1}{4}2^4-\frac{2}{3}2^3-(\frac{1}{4}0^4-\frac{2}{3}0^3)= -1,33 - 0 = -1,33\)

b)
\(\int_{0}^{3} {2e^{-3x+1}}\, dx = [-\frac{2}{3}e^{-3x+1}]_{0}^{3} = G(3)-G(0) = -\frac{2}{3}e^{-3\cdot 3+1}-(-\frac{2}{3}e^{-3\cdot 0+1}) = 1,81\)

c)
\(\int_{0}^{3} {h(x)}\, dx = [H(x)]_{0}^{3} = H(3)-H(0) = 4,13-3 = 1,13\)