Question

calcule a área delimitada pelo gráfico das funções y1= x² e y2= - x² + 8x

Answer 1

$pontos\,de\,\cap:$

$\mathsf{{x}^{2}=-{x}^{2}+8x}\\\mathsf{{x}^{2}+{x}^{2}-8x=0}\\\mathsf{2{x}^{2}-8x=0}\\\mathsf{2x(x-4)=0}$

$\mathsf{2x=0\rightarrow\,x=0}\\\mathsf{x-4=0\rightarrow\,x=4}$

$\mathsf{A_{(R)}=\int\limits_{0}^{4}(-{x}^{2}+8x-{x}^{2})dx}\\\mathsf{A_{(R)}=\int\limits_{0}^{4}(8x-2{x}^{2})dx}$

$\mathsf{A_{(R)}=4{x}^{2}-\dfrac{2}{3}.{x}^{3}\big|_{0}^{4}}\\\mathsf{A_{(R)}=4.{4}^{2}-\dfrac{2}{3}.{4}^{3}} \\\mathsf{A_{(R)}=64-\dfrac{128}{3}=\dfrac{192-128}{3}}$

$\boxed{\boxed{\mathsf{A_{(R)}=\dfrac{64}{3}u.a}}}$