Pair of Parabolas

The Area region is shown in the above graph.

The modified equations of the graphs are: $y = 4x - x^2 \ i.e., (x-2)^2 = -(y - 4) \ and \ y = x^2-x \ i.e., (x - \frac{1}{2})^2 = y + \frac{1}{4}$

$Required Area , A \ = \ Ar(OAEM) \ - \ Ar(DEM) \ + \ Ar(OCD)$ $= \int\limits_0^\frac{5}{2} (4x - x^2) dx - \int\limits_1^\frac{5}{2} (x^2 - x) dx - \int\limits_0^1 (x^2 - x) dx$ $\left[ { 2x^2 - \frac{x^3}{3} } \right] _{ 0 }^{ \frac{5}{2} } - \left[ {\frac{x^3}{3}-\frac{x^2}{2} } \right] _{ 1 }^{ \frac{5}{2} } - \left[ {\frac{x^3}{3}-\frac{x}{2} } \right] _{ 0 }^{ 1 }$ $= \left[2(\frac{25}{4}) - \frac{(\frac{5}{2})^3}{3}\right] - \left[ \frac{(\frac{5}{2})^3 - 1^3}{3} - \frac{(\frac{5}{2})^2 - 1^2}{2}\right] - \left[\frac{1^3}{3} - \frac{1}{2}\right]$ $= \frac{300 - 125 - 125 + 8 +75 -12 - 8 +12}{24}$ $= \frac{125}{24} sq. units$

$\Rightarrow 24A = 24(\frac{125}{24}) = \boxed{125 \ sq. units}$

Graphically, we clearly see that $A=(\int_0^{2.5} (4x-x^2) dx)-(\int_1^{2.5} (x^2-x) dx)-(\int_0^1 (x^2-x) dx)$

We then find $A=125/24$ . Hence, $24A=125/24*24 \boxed{=125}$