Question

Find the area bounded by the curve y = ${\displaystyle \sqrt{x}}$, x = 2y + 3 in the first quadrant and x-axis.

Answer 1

Given that: y = ${\displaystyle \sqrt{x}}$, x = 2y + 3, first quadrant and x-axis.

Solving y = ${\displaystyle \sqrt{x}}$ and x = 2y + 3

We get y = ${\displaystyle \sqrt{2y+3}}$

⇒ y² = 2y + 3

⇒ y² – 2y – 3 = 0

⇒ y² – 3y + y – 3 = 0

⇒ y(y – 3) + 1(y – 3) = 0

⇒ (y + 1)(y – 3) = 0

∴ y = –1, 3

Area of shaded region

= ${\displaystyle {\int }_0^3(2y+3)\text{d}y-{\int }_0^3{\text{y}}^2 \text{d}y}$

= ${\displaystyle {[2\frac{y^2}{2}+3y]}_0^3-\frac{1}{3}{\left[y^3\right]}_0^3}$

= ${\displaystyle [(9+9)-(0+0)]-\frac{1}{3}[27-0]}$

= 18 – 9

= 9 sq.units

Hence, the required area = 9 sq.units