The area of the region { ( x , y ) : x 2 + 4 x + 2 ≤ y ≤ ∣ x + 2∣ } is equal to

x 2 + 4 x + 2 ≤ y ≤ ∣ x + 2∣ A 1 = ∫ − 4 0 [ 2 − ( x 2 + 4 x + 2 ) ] d x − 2 1 × 4 × 2 = ( 3 − x 3 − 2 x 2 ) − 4 0 − 4 = 0 − ( 3 64 − 32 ) − 4 = 32 − 3 64 − 4 = 3 20

Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesHard2 minPYQ_2025

MathematicsHardsingle choice

The area of the region ${(x, y) : x^{2} + 4 x + 2 \leq y \leq ∣ x + 2∣}$ is equal to

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Answer:

Solution:

$x^{2} + 4 x + 2 \leq y \leq ∣ x + 2∣$ $A_{1} = \int_{- 4}^{0} [2 - (x^{2} + 4 x + 2)] d x - \frac{1}{2} \times 4 \times 2 = (\frac{- x ^{3}}{3} - 2 x^{2})_{- 4}^{0} - 4 = 0 - (\frac{64}{3} - 32) - 4 = 32 - \frac{64}{3} - 4 = \frac{20}{3}$

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

⏱ 2mℹ️ Source: PYQ_2025

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