The area of the region: R = x , y : 5 x 2 ≤ y ≤ 2 x 2 + 9 is

Required area = 2 ∫ 0 3 2 x 2 + 9 - 5 x 2 d x = 2 9 x - x 3 0 3 = 2 9 3 - 3 3 = 12 3 square units

Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesMedium2 minPYQ_2021

MathematicsMediumsingle choice

The area of the region: $R = \{(x, y) : 5 x^{2} \leq y \leq 2 x^{2} + 9\}$ is

Answer:

Solution:

Required area $= 2 \int_{0}^{\sqrt{3}} (2 x^{2} + 9 - 5 x^{2}) d x$

$= 2 {[9 x - x^{3}]}_{0}^{\sqrt{3}}$

$= 2 [9 \sqrt{3} - 3 \sqrt{3}] = 12 \sqrt{3}$ square units

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

⏱ 2mℹ️ Source: PYQ_2021

Loading discussions...