Gegeben

f (x) = e^{x} g (x) = - \frac{x}{2} + 1

M = {(x, y) ∣ 0 \leq y \leq min (f (x), g (x))}

Skizze

A = \int_{- \infty}^{0} e^{x} d x + \int_{0}^{2} - \frac{x}{2} + 1 d x

\int_{- \infty}^{0} e^{x} d x = r \to - \infty lim \int_{r}^{0} e^{x} d x = r \to - \infty lim [e^{x}]_{r}^{0} = r \to - \infty lim (e^{0} - e^{r}) = 1 - x \to \infty lim \frac{1}{e ^{x}} = 1 - 0 = 1

\int_{0}^{2} - \frac{x}{2} + 1 d x = [- \frac{1}{2} \cdot \frac{1}{2} x^{2} + x]_{0}^{2} = (- \frac{1}{4} \cdot 2^{2} + 2) - 0 = 1

A = 1 + 1 = \underline{2}