Calcular a área da seguinte figura usando integral.... Pergunta de ideia deMhary11

December 2019 1 85 Report

Calcular a área da seguinte figura usando integral

Lista de comentários

niltonjunior20oss764 $\mathrm{f(x)=x^2-5x+6\ \ \| \ \ g(x)=x}\\\\ \textbf{Abscissas de intersec\c{c}\~ao:}\\\\ \mathrm{f(x)=g(x)\ \to\ x^2-5x+6=x\ \to\ x^2-6x+6=0}\\\\ \mathrm{x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4.1.6}}{2.1}=\dfrac{6\pm\sqrt{36-24}}{2}=}\\\\ \mathrm{=\dfrac{6\pm\sqrt{12}}{2}=\dfrac{6\pm2\sqrt{3}}{2}=\boxed{\mathrm{3\pm\sqrt{3}}}}$

$\textbf{Calculando a \'area entre as curvas:}\\\\ \mathrm{\int_a^bf(x)-g(x)\ dx=\int_{3-\sqrt{3}}^{3+\sqrt{3}}x^2-5x+6-x\ dx=}\\\\ \mathrm{=\bigg(\int x^2-6x+6\ dx\bigg)\bigg|_{3-\sqrt{3}}^{3+\sqrt{3}}=\bigg(\int x^2\ dx-6\int x\ dx+\int6\ dx\bigg)\bigg|_{3-\sqrt{3}}^{3+\sqrt{3}}}=}\\\\\ \mathrm{=\bigg(\dfrac{x^3}{3}-6\dfrac{x^2}{2}+6x\bigg)\bigg|_{3-\sqrt{3}}^{3+\sqrt{3}}=\bigg(\dfrac{x^3-9x^2+18x}{3}\bigg)\bigg|_{3-\sqrt{3}}^{3+\sqrt{3}}=}$
$\mathrm{=\bigg(\dfrac{x(x^2-9x+18)}{3}\bigg)\bigg|_{3-\sqrt{3}}^{3+\sqrt{3}}=\bigg(\dfrac{x(x-6)(x-3)}{3}\bigg)\bigg|_{3-\sqrt{3}}^{3+\sqrt{3}}=}\\\\ \mathrm{=\bigg(\dfrac{(3+\sqrt{3})(\sqrt{3}-3)\sqrt{3}}{3}\bigg)-\bigg(\dfrac{(3-\sqrt{3})(-\sqrt{3}-3)(-\sqrt{3})}{3}\bigg)=}\\\\\ \mathrm{=\bigg(\dfrac{(3-9)\sqrt{3}}{3}\bigg)-\bigg(\dfrac{(9-3)\sqrt{3}}{3}\bigg)=-\dfrac{6\sqrt{3}}{3}-\dfrac{6\sqrt{3}}{3}=-4\sqrt{3}}\\\\\\ \mathrm{S=|-4\sqrt{3}|\ \to\ \boxed{\boxed{\mathbf{S=4\sqrt{3}\ u.a.}}}}$

0 votes Thanks 0