Question 148 of 230MediumMCQ
The area (in square units) bounded by the curves y=x, 2y−x+3=0, the x-axis, and lying in the first quadrant is: Solutions · 1
Official solution
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 \[ = \int\limits_0^3 {\left\{ {\left( {2y + 3} \right) - {y^2}} \right\}} dy\]
\[\left. {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {y^2} + 3y - {{{y^3}} \over 3}} \right|_0^3 = 9.\] Answer the question to unlock