Answer
$2.6667$
Work Step by Step
Here, the area can be expressed as: $A=\int_{-2}^{-1.69} [2^x-x-2] \ dx+\int_{-1.69}^{2} (x+2-2^x) \ dx$
In order to solve the above integral, we will use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Now, we have $\int_{-2}^{-1.69} [2^x-x-2] \ dx+\int_{-1.69}^{2} (x+2-2^x) \ dx=[\dfrac{2^x}{\ln 2}-\dfrac{x^2}{2}-2x]_{-2}^{-1.69}+[\dfrac{x^2}{2}+2x-\dfrac{2^x}{\ln 2}]_{-1.69}^{2}$
or, $=[\dfrac{2^{-1.69}}{\ln 2}-\dfrac{(-1.69)^2}{2}-2(-1.69)]-[\dfrac{2^{-2}}{\ln 2}-\dfrac{(-2)^2}{2}-2(-2)]+[\dfrac{2^2}{2}+2(2)-\dfrac{2^{2}}{\ln 2}]-[\dfrac{(-1.69)^2}{2}+2(-1.69)-\dfrac{2^{-1.69}}{\ln 2}] $
Therefore, the required area is: $Area \approx 2.6667$
