QJEEM13Ma62ac: JEE Main 2013 Mathematics - Class 12 JEEM13Ma62ac Solution: The area (in square units) bounded by the curves -axis, and…

Question

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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 :

Accepted Answer

Given curves are

y = \sqrt x

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

and

2y - x + 3 = 0\,\,\,\,\,\,\,\,\,...\left( 2 \right)

On solving both we get

y=-1,3

Required area

= \int\limits_0^3 {\left\{ {\left( {2y + 3} \right) - {y^2}} \right\}} dy

\left. {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {y^2} + 3y - {{{y^3}} \over 3}} \right|_0^3 = 9.

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