Find the area bounded by the curves $x^{2} = y, x^{2} = - y$ and $y^{2} = 4 x - 3$

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Solution

The region bounded by the given curves
$x^{2} = y, x^{2} = - y$ and $y^{2} = 4 x - 3$ symmetrical about the xaxis.
The parabolas $x^{2} = y$ and $y^{2} = 4 x - 3$ touch at the point (1, 1).
Moreover the vertex of the curve $y^{2} = 4 x - 3$ is at $(\frac{3}{4}, 0)$
Hence the area of the region $= 2 [\int_{0}^{1} x^{2} d x - \int_{3 / 4}^{1} \sqrt{4 x - 3} d x]$
$= 2 [(\frac{x^{3}}{3})_{0}^{1} - \frac{1}{6} ((4 x - 3)^{3 / 2})_{3 / 4}^{1}] = 2 [\frac{1}{3} - \frac{1}{6}] = \frac{1}{3}$ . sq. units
The region bounded by the curves $y = x^{2}, y = - x^{2}$ and $y^{2} = 4 x - 3$ meets at $(1, 1)$
Area at curve $(O A B C O) = 2 [\int_{0}^{1} x^{2} d x - \int_{\frac{3}{4}}^{1} (\sqrt{4 x - 3}) d x]$
$= 2 ⎡ ⎢ ⎢ ⎣ {(\frac{x^{3}}{3})}_{0}^{1} - {⎛ ⎜ ⎝ \frac{{(4 x - 3)}^{\frac{3}{2}}}{3. \frac{4}{2}} ⎞ ⎟ ⎠}_{\frac{1}{4}}^{1} ⎤ ⎥ ⎥ ⎦$
$= 2 (\frac{1}{3} - \frac{1}{6}) = \frac{1}{3}$ units

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