Question

utilizando integrais dupla calcule a area limitada pelas funcoes y=3 e y=x^2-4x+4

Answer 1

$\bmatrix y_a=3\\y_b=x^2-4x+4\end$

encontrando o intervalo

$y_a=y_b\\\\3=x^2-4x+4\\0=x^2-4x+1 \to \bmatrix x_1=2-\sqrt{3}\\x_2=2+\sqrt{3}\end$


observando o grafico vemos que a area procurada é

$\boxed{\boxed{A= \int\limits^{2+\sqrt{3}}_{2-\sqrt{3}} \int^{3}_{x^2-4x+4} \; {dy\;dx}}} \\\\\\ A= \int\limits^{2+\sqrt{3}}_{2-\sqrt{3}} \left(3-(x^2-4x+4)\right)dx\\\\\\ A= \int\limits^{2+\sqrt{3}}_{2-\sqrt{3}} \left(-x^2+4x-1\right)dx\\\\\\A = \left -\frac{x^3}{3}+ \frac{4x^2}{2}-x \right |^{2+\sqrt{3}}_{2-\sqrt{3}} \\\\\\ A = \left -\frac{x^3}{3}+2x^2-x \right |^{2+\sqrt{3}}_{2-\sqrt{3}}$


$A= \scriptscriptstyle \left( -\frac{(2+\sqrt{3})^3}{3}+2(2+\sqrt{3})^2-(2+\sqrt{3}) \right) - \left( -\frac{(2-\sqrt{3})^3}{3}+2(2-\sqrt{3})^2-(2-\sqrt{3}) \right) \\\\A=4 \sqrt{3} \approx 6,93 \; u.a$