The area bounded by the curve $y^{2} = 4 x$ and line $x = 1$

\frac{8}{3}

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\frac{7}{3}

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\frac{4}{3}

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\frac{9}{2}

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Solution

The correct option is A $\frac{8}{3}$
Due to symmetry
area above X-axis
$2 \times 1 - \int_{0}^{2} \frac{y^{2}}{4} d y$
$= 2 - [{\frac{y^{3} 12}{]}}_{0}^{2}$
$= 2 - \frac{8}{12} = \frac{4}{3}$
total area $= \frac{8}{3}$

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