The parabolas y^2 = 4x and x^2 = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes into three parts. If S_1, S_2, - Mathematics MCQ AIEEE 2026 | ExamTest.live

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The parabolas y^{2} = 4 x and x^{2} = 4 y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes into three parts. If S_{1}, S_{2}, S_{3} are respectively the areas of these parts numbered from top to bottom, then S_{1} : S_{2} : S_{3} is:

Solution & Step-by-step Explanation

The square area is 4 \times 4 = 16 .Area S_{2} bounded by the two parabolas: S_{2} = \int_{0}^{4} (4 x - \frac{x ^{2}}{4}) d x = [\frac{2 \cdot 2 x ^{3/2}}{3} - \frac{x ^{3}}{12}]_{0}^{4} S_{2} = \frac{4 \cdot 8}{3} - \frac{64}{12} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} Area S_{3} below x^{2} = 4 y : S_{3} = \int_{0}^{4} \frac{x ^{2}}{4} d x = [\frac{x ^{3}}{12}]_{0}^{4} = \frac{64}{12} = \frac{16}{3} Area S_{1} above y^{2} = 4 x :Total area = S_{1} + S_{2} + S_{3} = 16 . S_{1} = 16 - (16/3 + 16/3) = 16 - 32/3 = 16/3 .Thus, S_{1} : S_{2} : S_{3} = \frac{16}{3} : \frac{16}{3} : \frac{16}{3} = 1 : 1 : 1 .

Try it yourself before checking the explanation above.

The parabolas y^{2} = 4 x and x^{2} = 4 y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes into three parts. If S_{1}, S_{2}, S_{3} are respectively the areas of these parts numbered from top to bottom, then S_{1} : S_{2} : S_{3} is:

1 : 2 : 1

1 : 2 : 3

2 : 1 : 2

1 : 1 : 1

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