Transformation Complication

First, I think the picture isn't quite right in that it was horizontally reflected.

Ok, let's get to the solution.

Step 1: Find the equation of the transformed graph $x'=2x+y$ $y'=3x+2y$

The original equation: $y=x^2$

The transformed graph: $3x+2y=(2x+y)^2$

Step 2: Solve for points of intersection between these two graphs via simultaneous equations $y=x^2$ $3x+2y=(2x+y)^2$ Which is relatively straightforward, giving us 4 coordinates: $(0,0),(-3,9),(\frac{-1+\sqrt(5)}{2},\frac{3-\sqrt(5)}{2}),(\frac{-1-\sqrt(5)}{2},\frac{3+\sqrt(5)}{2})$

Step 3: Make y the subject in the equation of the transformed graph $y=-2x\pm \sqrt{1-x}+1$

Step 4: Find multiples areas, which will be added/subtracted with each other to find the final area

Let the final area be X

Let A,B,C,D be these areas $A=\int_{-3}^{\frac{-1-\sqrt{5}}{2}} \!x^2\, \mathrm{d}x$ $B=\int_{\frac{-1-\sqrt{5}}{2}}^{0} \!-2x-\sqrt{1-x}+1\, \mathrm{d}x$ --> chose minus because looking at the lower curve $C=\int_{0}^{\frac{-1+\sqrt{5}}{2}} \!x^2\, \mathrm{d}x$ $D=\int_{-3}^{\frac{-1+\sqrt{5}}{2}} \!-2x+\sqrt{1-x}+1\, \mathrm{d}x$ --> chose plus because looking at the upper curve

Evaluating all 4 of these, we get $A=\frac{25-\sqrt{5}}{3},B=\frac{4+\sqrt{5}}{3},C=\frac{-2+\sqrt{5}}{3},D=\frac{50+\sqrt{5}}{3}$

Step 5: Calculate the final area $X=D-A-B-C=\frac{23}{3}$