The area (in sq. units) of the region x , y ∈ R 2 4 x 2 ≤ y ≤ 8 x + 12 is

4 x 2 = y y = 8 x + 12 4 x 2 = 8 x + 12 x 2 - 2 x - 3 = 0 x = - 1 , 3 A = ∫ - 1 3 8 x + 12 - 4 x 2 d x A = 8 x 2 2 + 12 x - 4 x 3 3 3 - 1 = 4 9 + 36 - 36 - 4 - 12 + 4 3 = 36 + 8 - 4 3 = 44 - 4 3 = 132 - 4 3 = 128 3

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Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesMedium2 minPYQ_2020

MathematicsMediumsingle choice

The area (in sq. units) of the region $\{(x, y) \in R^{2} |4 x^{2} \leq y \leq 8 x + 12\}$ is

Options:

Answer:

Solution:

$4 x^{2} = y$
$y = 8 x + 12$
$4 x^{2} = 8 x + 12$
$x^{2} - 2 x - 3 = 0$
$x = - 1, 3$
$A = \int_{- 1}^{3} (8 x + 12 - 4 x^{2}) d x$
$A = \frac{8 x^{2}}{2} + 12 x - \frac{4 x^{3}}{3} \begin{matrix} 3 \\ - 1 \end{matrix} = (4 (9) + 36 - 36) - (4 - 12 + \frac{4}{3}) = 36 + 8 - \frac{4}{3}$
$= 44 - \frac{4}{3} = \frac{132 - 4}{3} = \frac{128}{3}$

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

⏱ 2mℹ️ Source: PYQ_2020

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