Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesEasy2 minPYQ_2022

MathematicsEasysingle choice

The area of the region bounded by $y^{2} = 8 x$ and $y^{2} = 16 (3 - x)$ is equal to

Answer:

Solution:

Given curves are $y^{2} = 8 x$ and $y^{2} = 16 (3 - x)$

$\Rightarrow 8 x = 48 - 16 x$ $\Rightarrow x = 2$ , so $y = \pm 4$

Required area $= 2 \int_{0}^{4} ((3 - \frac{y^{2}}{16}) - \frac{y^{2}}{8}) d y$

$= 2 {[3 y - \frac{y^{3}}{48} - \frac{y^{3}}{24}]}_{0}^{4}$ $= 2 [12 - \frac{64}{48} - \frac{64}{24}] = 16$

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

⏱ 2mℹ️ Source: PYQ_2022

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