Mathematics - Area Under Curves Question with Solution | TestHub

The given parabola is $y^2=4ax$.

The area bounded by the parabola and the lines $x=a$ and $x=4a$ is symmetric about the x-axis.

Required area $A=2{\int }_a^{4a}ydx$

$A=2{\int }_a^{4a}\sqrt{4ax}dx$

$A=4\sqrt{a}{\int }_a^{4a}{x}^{1/2}dx$

$A=4\sqrt{a}{\left[\frac{x^{3/2}}{3/2}\right]}_a^{4a}$

$A=\frac{8\sqrt{a}}{3}\left[(4a{)}^{3/2}-(a{)}^{3/2}\right]$

$A=\frac{8\sqrt{a}}{3}\left[8a\sqrt{a}-a\sqrt{a}\right]$

$A=\frac{8\sqrt{a}}{3}\left[7a\sqrt{a}\right]$

$A=\frac{56a^2}{3}$ sq. units.

Answer: $\frac{56a^2}{3}$ sq. units