Mathematics - Area Under Curves Question with Solution | TestHub

The given equations of the parabolas are $4y^2=9x$ and $3x^2=16y$.

From the second equation, we have $y=\frac{3x^2}{16}$. Substituting this into the first equation:

$4{\left(\frac{3x^2}{16}\right)}^2=9x$

$\frac{9x^4}{64}=9x$

$x^4-64x=0$

$x(x^3-64)=0$

The real roots are $x=0$ and $x=4$.

The required area $A$ is the integral of the upper curve minus the lower curve from $x=0$ to $x=4$:

$A={\int }_0^4\left(\frac{3}{2}\sqrt{x}-\frac{3x^2}{16}\right)dx$

$A={\left[x^{3/2}-\frac{x^3}{16}\right]}_0^4$

$A=\left(4^{3/2}-\frac{4^3}{16}\right)-0$

$A=8-4=4$ sq. units.

Answer: 4 sq. units