JEE MAIN - Mathematics (2013 (Offline) - No. 5)
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 :
$$9$$
$$36$$
$$18$$
$${{27} \over 4}$$
Explanation
Given curves are
$$y = \sqrt x $$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
and $$2y - x + 3 = 0\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
On solving both we get $$y=-1,3$$
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Required area $$ = \int\limits_0^3 {\left\{ {\left( {2y + 3} \right) - {y^2}} \right\}} dy$$
$$\left. {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {y^2} + 3y - {{{y^3}} \over 3}} \right|_0^3 = 9.$$
$$y = \sqrt x $$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
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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