Mathematics - Area Under Curves Question with Solution | TestHub

The given equation of the parabola is $y^2=27x$.

The region is bounded by the parabola and the vertical line $x=1$. Since the parabola is symmetric about the x-axis, the required area is twice the area in the first quadrant.

Area $A=2{\int }_0^{1}ydx$

Substituting $y=\sqrt{27x}=3\sqrt{3}\sqrt{x}$:

$A=2{\int }_0^{1}3\sqrt{3}\sqrt{x}dx$

$A=6\sqrt{3}{\int }_0^1{x}^{1/2}dx$

$A=6\sqrt{3}{\left[\frac{x^{3/2}}{3/2}\right]}_0^1$

$A=6\sqrt{3}\times \frac{2}{3}(1-0)$

$A=4\sqrt{3}$ sq. units.

Answer: $4\sqrt{3}$