I think that you mean the area bound ONLY by the parabola an the straight line.
Part of this area will have a negative sign; therefore you have to find the area in pieces.
The area enclosed by the parabola and the x-axis is twice the area bounded by the parabola, the y-axis and the x-axis to the right of the origin, i.e. twice the integral from 0 to 2 of 4 - x2; that area is 32/3
The triangular area bounded by the straight line and x = 1 is 9/2.
the area bounded by the parabola, x = 1 and the x-axis is the integral of 4 - x2 between 1 and 2; this area is 5/3.
The area required then is (32/3) - (9/2) - (5/3) = 9/2.