Question

Find the area of the region enclosed by the parabola $y^2=5x+6$ and $x^2=y$.

Answer 1

Area enclose by $y^2=5x+6,$

and $x^2=y$

$\Rightarrow x^4-1=5x+5(x^2+1)(x-1)(x+1)-5(x+1)=0x=-1$

or, $(x^2+1)(x-1)-5=0x^3-x^2=x-1-5=0x^3-x^2+6=0x^3-2x^2+x^2-2x+3x-6=0x^3(x-2)+x(x-2)+3(x-2)=0(x-2)(x^2+x+3)=0x-2$

or $x^2+x+3=0$

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

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

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

$=\frac{2}{11}[{\left(16\right)}^{\frac{3}{2}}-1^{\frac{3}{2}}]-[\frac{8}{3}+\frac{1}{3}]=\frac{2}{11}[4^3-1]-\frac{9}{3}=\frac{2}{11}\times 63-3=\frac{126}{15}-3=\frac{42-15}{5}=\frac{27}{5}squnit$