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JAMIA MILLIA ISLAMIA MCA 2024

The area bounded by the curves $y^2 = 4x$ and $y = x$ is equal to

Solution

For intersection: $y^2 = 4x$ and $y = x \Rightarrow x^2 = 4x \Rightarrow x = 0, 4$ Area $= \int_{0}^{4} (\sqrt{4x} - x),dx = \int_{0}^{4} (2\sqrt{x} - x),dx = \left[\dfrac{4}{3}x^{3/2} - \dfrac{x^2}{2}\right]_{0}^{4} = \dfrac{32}{3} - 8 = \dfrac{8}{3}$
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