Mathematics - Area Under Curves Question with Solution | TestHub

The given curves are $y=\sqrt{x}$ and $2y-x+3=0$.

To find the point of intersection, substitute $x=y^2$ into the equation of the line:

$2y-y^2+3=0$

$y^2-2y-3=0$

$(y-3)(y+1)=0$

Since the region lies in the first quadrant, $y\ge 0$, which gives $y=3$. The corresponding $x$-coordinate is $x=9$.

The line $2y-x+3=0$ intersects the X-axis at $x=3$.

The required area can be found by integrating with respect to $y$ from $y=0$ to $y=3$. The right boundary is the line $x=2y+3$ and the left boundary is the curve $x=y^2$.

Required Area $={\int }_0^3((2y+3)-y^2)dy$

$={\left[y^2+3y-\frac{y^3}{3}\right]}_0^3$

$=\left(3^2+3(3)-\frac{3^3}{3}\right)-0$

$=9+9-9=9$ sq. units.

Answer: 9