• # question_answer The area bounded by the curves $y=\sqrt{x},$ $2y+3=x$ and $x-$axis in the 1st quadrant is                    [IIT Screening 2003] A)            9     B)            $\frac{27}{4}$ C)            36   D)            18

Solving ${{y}^{2}}=x$and $x=2y+3$                    $4{{y}^{2}}={{(x-3)}^{2}}$, $4x={{x}^{2}}-6x+9$                    Þ ${{x}^{2}}-10x+9=0$Þ $(x-1)(x-9)=0$Þ $x=1,\,9$        =$-\,4\,[x\log x-x]_{0}^{1}=-4(-1)=4$sq. unit,  $(\because \text{ }\underset{\text{x}\to \text{0}}{\mathop{\text{lim}}}\,x\log x=0)$.       Required area = A+B$=\int_{0}^{3}{\sqrt{x}dx+\int_{3}^{9}{\left[ \sqrt{x}-\left( \frac{x-3}{2} \right) \right]}}dx$            $=\frac{2}{3}[{{x}^{3/2}}]_{0}^{3}+\frac{2}{3}[{{x}^{3/2}}]_{3}^{9}-\frac{1}{2}\left[ \frac{{{x}^{2}}}{2}-3x \right]_{3}^{9}$            $=\frac{2}{3}3\sqrt{3}+\frac{2}{3}[9\times 3-3\sqrt{3}]-\frac{1}{2}\left[ \left( \frac{81}{2}-27 \right)-\left( \frac{9}{2}-9 \right) \right]$            $=18-\frac{1}{2}[36-18]=18-9=9$ sq. unit.