Find the area of the region bounded by the curves $y = \sqrt{x}$, $2y - x + 3 = 0$, $x$-axis and lying in the first quadrant.

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> [!TLDR]
> The curve $y=\sqrt{x}$ (i.e. $x=y^2$) and the line $x=2y+3$ meet at $(9,3)$; integrating $x_{	ext{line}}-x_{	ext{curve}}$ in $y$ from $0$ to $3$ gives an area of $9$ square units.
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> **Concept.** For a region more naturally described as lying between a left curve and a right curve, integrate the horizontal width $(x_{	ext{right}}-x_{	ext{left}})$ with respect to $y$. This applications-of-integrals technique is standard in NCERT Class 12 mathematics.
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> **Boundaries.** $y=\sqrt{x}\Rightarrow x=y^2$. The line $2y-x+3=0\Rightarrow x=2y+3$; it meets the $x$-axis ($y=0$) at $x=3$.
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> **Intersection of curve and line.** $\sqrt{x}=\frac{x-3}{2}\Rightarrow 2\sqrt{x}=x-3$. Put $t=\sqrt{x}$: $t^2-2t-3=0\Rightarrow(t-3)(t+1)=0\Rightarrow t=3$, so $x=9,\ y=3$.
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Q.Find the area of the region bounded by the curves $y = \sqrt{x}$ , $2y - x + 3 = 0$ , $x$ -axis and lying in the first quadrant.

Concept understanding — Area between Two Curves

Q.Find the area of the region bounded by the curves y=xy = \sqrt{x}y=x​, 2y−x+3=02y - x + 3 = 02y−x+3=0, xxx-axis and lying in the first quadrant.

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Q.Find the area of the region bounded by the curves $y = \sqrt{x}$ , $2y - x + 3 = 0$ , $x$ -axis and lying in the first quadrant.

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