6.1 Areas Between Curves/20

$4x+y^{2}=12,x=y$

${\frac {4x}{4}}={\frac {-y^{2}+12}{4}}$

$x=-{\frac {1}{4}}y^{2}+3$

$\int _{0}^{3}(-{\frac {1}{4}}y^{2}-y+3)-y$

$\int _{0}^{3}-{\frac {1}{4}}y^{2}-y+3$

${\frac {{\frac {1}{4}}y^{3}}{3}}-{\frac {y^{2}}{2}}+3y$

$-{\frac {1}{12}}y^{3}-{\frac {1}{2}}y^{2}+3y{\bigg |}_{0}^{3}$

$-{\frac {1}{12}}(3)^{3}-{\frac {1}{2}}(3)^{2}+3(3)$

$=2.25$