Answer
$0.9138$
Work Step by Step
Here, the area can be expressed as: $A=\int_{-1}^{0} [e^x-(2x+1)] \ dx+\int_{0}^{1} [(2x+1)-e^x] \ dx=\int_{-1}^{0} (e^x-2x-1) \ dx+\int_{0}^{1} (2x+1-e^x) \ dx$
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 $\int_{-1}^{0} (e^x-2x-1) \ dx+\int_{0}^{1} (2x+1-e^x) \ dx=[e^x -x^2 -x]_{-1}^0+[x^2 +x -e^x]_0^1$
or, $=1-(\dfrac{1}{e}-1+1)+(1+1-e)+1$
or, $=4-\dfrac{1}{e}-e$
Therefore, the required area is: $Area \approx 0.9138$
