Application of Integrals: The area (in square units) bounded by the curves y=sqrt x , 2y-x+3=0… | JEE Main 2013 | Yukti Vault

Yukti Vault

JEE Main 2013 (Offline)Question 1

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 :

Application of Integrals

$9$✓ Correct answer$36$$18$$\frac{27}{4}$

Solution

$y=\sqrt x$ meets $x=2y+3$ at $(9,3)$. Area $=\int_0^3(2y+3-y^2)dy=9+9-9=9$.

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