Question

The area bounded by the parabola $y^2=5x+6$ and $x^2=y$

Answer 1

REF.Image

$y^2=x^4=5x+6$

$x^4-5x+6=0$

$x=-1$ $x=2$

Area = ${\int }_{-1}^2(\sqrt{5x+6}-x^2)dx$

$=[\frac{2(5x+6{)}^{3/2}}{3\times 5}-\frac{x^3}{3}{]}_{-1}^2$

$\left(\frac{2}{15}\right(5\left(2\right)+6{)}^{3/2}-\frac{8}{3})-(\frac{2}{15}\left(5\right(-1)+6)+\frac{1}{3})$

$\frac{2}{15}{4}^{2\times \frac{3}{2}}-\frac{8}{3}-\frac{2}{15}-\frac{1}{3}$

$\frac{128-2}{15}-\frac{8-1}{3}$

$\frac{126}{15}-\frac{9}{3}\Rightarrow \frac{126-27}{15}$

$\Rightarrow \frac{99}{15}=\frac{33}{5}$