CUET · MATHS · PYQ PAPER 2025
Area of the region bounded by the curve \(y=\sqrt{x}\) and lines \(x+y=2, y=0\) is
- A \(\frac{9}{6}\) square units
- B \(\frac{9}{2}\) square units
- C \(\frac{7}{6}\) square units
- D \(\frac{7}{3}\) square units
Answer & Solution
Correct Answer
(C) \(\frac{7}{6}\) square units
Step-by-step Solution
Detailed explanation
Intersection of \(y=\sqrt{x}\) and \(x+y=2\): \(\sqrt{x}=2-x \implies x=1\), so point is \((1,1)\). Area \(A = \int_{0}^{1} ((2-y) - y^2) \, dy\) \(A = \left[ 2y - \frac{y^2}{2} - \frac{y^3}{3} \right]_{0}^{1}\) \(A = \left( 2(1) - \frac{1^2}{2} - \frac{1^3}{3} \right) - 0\)…
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