Area of region bounded by the curves \( y=\sqrt{x},\ x-2y=3 \) that lies in first quadrant is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 9 sq.units
The curves meet where \( x=y^{2} \) and \( x=2y+3 \):
\( y^{2}-2y-3=0 \)
\( (y-3)(y+1)=0 \)
Since \( y=\sqrt{x}\ge 0 \), only \( y=3 \) is possible, and the curves meet at \( (9,3) \).
In the first quadrant the parabola \( x=y^{2} \) lies to the left of the line \( x=2y+3 \). Integrating with respect to \( y \):
\( \displaystyle \text{Area}=\int_{0}^{3}\left[(2y+3)-y^{2}\right]dy \)
\( \displaystyle =\left[y^{2}+3y-\frac{y^{3}}{3}\right]_{0}^{3} \)
\( =9+9-9 \)
\( =9 \)
Tip: The area under the parabola alone, \( \int_{0}^{9}\sqrt{x}\,dx=18 \), is Option (2). The triangle between the line and the x-axis (area 9) has to be removed from it.
NCERT Exemplar:
Application of Integrals
