Question

Encontre a área delimitada pelas funções y=5x-x2 e y=x e descreva todo o processo. A imagem da área está representada na figura abaixo.

Answer 1

$\large\boxed{\begin{array}{l}\underline{\sf pontos\,de\,intersecc_{\!\!,}\tilde ao:}\\\rm x=5x-x^2\\\rm x^2-5x+x=0\\\rm x^2-4x=0\\\rm x\cdot(x-4)=0\\\rm x=0\\\rm x-4=0\\\rm x=4\\\rm se~x=0\longrightarrow y=0\\\rm se~x=4\longrightarrow y=4\\\rm A(0,0)~~~B(4,4)\end{array}}$

$\large\boxed{\begin{array}{l}\rm a~\acute area~pedida~\acute e~dada~por\\\displaystyle\rm\int_0^4(5x-x^2-[x])\,dx\\\displaystyle\rm\int_0^4(5x-x^2-x)\,dx\\\displaystyle\rm\int_0^4(4x-x^2)\,dx=4\cdot\dfrac{x^{1+1}}{1+1}-\dfrac{x^{2+1}}{2+1}\\\\\displaystyle\rm\int _0^4(4x-x^2)=\bigg[ 2x^2-\dfrac{1}{3}x^3\bigg]_0^4\\\sf Aqui\,n\tilde ao\,precisa\,substituir\,x\,por\,zero\\\sf pois\,tudo\,se\,anula.\\\sf desse ~modo:\\\displaystyle\rm\int_0^4(4x-x^2)\,dx=2\cdot4^2-\dfrac{1}{3}\cdot4^3\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\int_0^4(4x-x^2)\,dx=32-\dfrac{64}{3}\\\displaystyle\rm\int_0^4(4x-x^2)\,dx=\dfrac{32}{3}\,u\bullet a\end{array}}$