The area (in square units) bounded by the curves y = √(x,) 2y - x + 3 = 0, x -axis, and lying in the first quadrant is:

Integral Calculus: JEE Main 2013 Mathematics question

JEE Main 2013 · Session, Shift 1 · Q2802

JEE Main 2013Session, Shift 1MathematicsMediumIntegral Calculus369× since 2002

The area (in square units) bounded by the curves

y = \sqrt {x,}

2y - x + 3 = 0,

x

-axis, and lying in the first quadrant is :

JEE Main 2013 · Session, Shift 1 · Q2802

JEE Main 2013Session, Shift 1MathematicsMediumIntegral Calculus369× since 2002

The area (in square units) bounded by the curves

y = \sqrt {x,}

2y - x + 3 = 0,

x

-axis, and lying in the first quadrant is :