${\displaystyle g(x)=f(x)\rightarrow 0{,}5e^{x}-1\ =-0{,}16x^{2}-0{,}4x+2{,}75\ \rightarrow }$
${\displaystyle x_{1}\approx 1{,}66433}$

${\displaystyle A=\int _{-2}^{x_{1}}{\big (}f(x)-g(x){\big )}\,\mathrm {d} x+\int _{x_{1}}^{2}{\big (}g(x)-f(x){\big )}\,\mathrm {d} x=}$
${\displaystyle \int _{-2}^{x_{1}}(-0{,}5e^{x}+1-0{,}16x^{2}-0{,}4x+2{,}75)\,\mathrm {d} x+\dots }$
${\displaystyle \qquad \quad \dots +\int _{x_{1}}^{2}(0{,}5e^{x}-1+0{,}16x^{2}+0{,}4x-2{,}75)\,\mathrm {d} x=}$

${\displaystyle \left[-0{,}5e^{x}+x-{\frac {0{,}16}{3}}x^{3}-0{,}2x+2{,}75x\right]_{-2}^{x_{1}}+}$
${\displaystyle \qquad \quad \dots +\left[0{,}5e^{x}-x+{\frac {0{,}16}{3}}x^{3}+0{,}2x^{2}-2{,}75x\right]_{x_{1}}^{2}=}$

${\displaystyle \left(-0{,}5e^{1{,}66..}+1{,}66..-{\frac {0{,}16}{3}}\cdot {1{,}66..}^{3}-0{,}2\cdot {1{,}66..}^{2}+2{,}75\cdot 1{,}66..\right)-\dots }$
${\displaystyle \qquad \quad \dots -\left(-0{,}5e^{(-2)}+1-{\frac {0{,}16}{3}}\cdot {(-2)}^{3}-0{,}2\cdot {(-2)}^{2}+2{,}75\cdot (-2)\right)+\ {\text{usw.}}\ldots }$
${\displaystyle \ \approx 10{,}96\ }$ (mit Geogebra berechnet)