Finding the area of the part of the plane given by the curves I am supposed to find the area of the part of the plane given by the curves: $$y=x^{2},y=-x^{2}, -4x+4=y$$ I know that I have to divide it into two parts and so on.. but my question is.. how do I know which function to put first and second in the formula: $$\int _{a}^{b}(f(x)-g(x))dx$$ My idea for the first integral is : $$\int _{0}^{-2+2\sqrt{2}}(x^{2}+4x-4)dx$$ Is that correct?

**Hint** :

You can split into the area _above_ the $x$-axis, which involves only the upper parabola and the straight line and the area _below_ the $x$-axis, which involves the lower parabola and the line.

For each area, you can determine the intersection $A, B$ of the line with the upper and lower parabola respectively, and its $x$-intercept $C=(x_0,0)$. If $\xi_+$ and $\xi_-$ are the abscissæ of $A$ and $B$, $H,K$ their projections on the $x$-axis, the upper and lower areas are (see figure below): $$\mathcal A_+=\int_0^{\xi_+}\\!\\!x^2\,\mathrm dx+\mathcal A(\text{triangle }AHC),\qquad\mathcal A_-=\int^0_{\xi_-}\\!\\!x^2\,\mathrm dx-\mathcal A(\text{triangle }BKC). $$

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