Area bounded by the curve S 3 , y ≤ − 1 and the line x = 3 is

The line x = 3 meets S 3 at y = ± 3 . For − 3 ≤ y ≤ − 1 the region lies between x = 4 y 2 + 3 and x = 3 . Area = ∫ − 3 − 1 ( 3 − 4 y 2 + 3 ) d y = 6 − [ 12 y 3 + 4 3 y ] − 3 − 1 = 6 − 3 11 = 3 7 .

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Mathematics - Area Under Curves Question with Solution | TestHub

MathematicsArea Under CurvesArea bounded by two curvesMedium2 minQB

MathematicsMediumsingle choice

Passage / Comprehension

Let the curves be $S_{1} : y = x^{2}, S_{2} : y = - x^{2}, S_{3} : y^{2} = 4 x - 3$ .

Area bounded by the curve $S_{3}, y \leq - 1$ and the line $x = 3$ is

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Answer:

Solution:

The line $x = 3$ meets $S_{3}$ at $y = \pm 3$ . For $- 3 \leq y \leq - 1$ the region lies between $x = \frac{y ^{2} + 3}{4}$ and $x = 3$ . Area $= \int_{- 3}^{- 1} (3 - \frac{y ^{2} + 3}{4}) d y = 6 - [\frac{y ^{3}}{12} + \frac{3 y}{4}]_{- 3}^{- 1} = 6 - \frac{11}{3} = \frac{7}{3}$ .

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

⏱ 2mℹ️ Source: QB

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