The area bounded by the parabola $y^2=4 x$ and the line $2 x-3 y+4=0$, in square unit, is

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Aicharya · Accepted Answer

Intersecting points are $x=1,4$ $ \begin{aligned} & 	herefore 	ext { Required area }=\int_1^4\left[2 \sqrt{x}-\left(\frac{2 x+4}{3}ight)ight] d x \ & =\left.\frac{2 x^{3 / 2}}{3 / 2}ight|_1 ^4-\left.\frac{2 x^2}{3 	imes 2}ight|_1 ^4-\left.\frac{4}{3} xight|_1 ^4 \end{aligned} $ $ \begin{aligned} & =\frac{4}{3}\left(4^{3 / 2}-1^{3 / 2}ight)-\frac{1}{3}(16-1)\left[\frac{4}{3}(4)-\frac{4}{3}ight] \ & =\frac{4}{3}(7)-5-4=\frac{28}{3}-9=\frac{28-27}{3}=\frac{1}{3} \end{aligned} $

The area bounded by the parabola $y^2=4 x$ and the line $2 x-3 y+4=0$, in square unit, is

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