The area of the indicated region is 20/3 unit^2
Data;
To find the area under the curve, we have to integrate through the sides
[tex]A = \int\limits^x^=^1_x_=_-_1 [{\int\limits^x^2_y_=_-_3 {x} \, dy } \,] dx[/tex]
This becomes
[tex]A = \int\limits^x^=^1_x_=_-_1 {(x^2+3)} \, dx[/tex]
resolving this,
[tex]A = \frac{20}{3}[/tex]
The area of the indicated region is 20/3 unit^2
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