Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesMedium2 minPYQ_2025

MathematicsMediumsingle choice

The area of the region enclosed by the curves $y = x^{2} - 4 x + 4$ and $y^{2} = 16 - 8 x$ is :

Answer:

Solution:

$Area = \int_{0}^{2} (16 - 8 x - (x^{2} - 4 x + 4)) d x = \frac{- ( 16 - 8 x ) ^{3/2}}{12} - \frac{x ^{3}}{3} + 2 x^{2} + 4 x]_{0}^{2} = \frac{8}{3}$

Stream:JEESubject:MathematicsTopic:Area Under CurvesSubtopic:Area bounded by two curves

⏱ 2mℹ️ Source: PYQ_2025

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