Question

Siga as seguintes instruções para resolução do grafico das integrais

Answer 1

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$\sf ponto~de~intersecc_{\!\!,}\tilde ao~das~func_{\!\!,}\tilde oes:\\\sf x^2+2=-2x+10\\\sf x^2+2x+1+1=10\\\sf (x+1)^2=10-1\\\sf (x+1)^2=9\\\sf x+1=\pm\sqrt{9}\\\sf x+1=\pm3\\\sf x+1=3\implies x=2\\\sf x+1=-3\implies x=-4\\\sf se~x=2\implies y=2^2+2=6\\\sf se~x=-4\implies y=(-4)^2+2=18$

$\tt A_1=\displaystyle\sf\int_0^2(x^2+2)~dx=\dfrac{1}{3}x^3+2x\Bigg|_{0}^2\\\sf =\dfrac{1}{3}\cdot2^3+2\cdot2=\dfrac{8}{3}+4=\dfrac{20}{3}$

$\sf ra\acute iz~de~g(x)=-2x+10:\\\sf x=-\dfrac{10}{-2}=5\\\tt A_2=\displaystyle\sf\int_2^5(-2x+10)~dx=-x^2+10x\Bigg|_{2}^5\\\sf=-5^2+10\cdot5-(-2^2+10\cdot2)\\\sf =-25+50+4-20=9$

$\rm A_{figura}=\tt A_1+\tt A_2\\\rm A_{figura}=\dfrac{20}{3}+9\\\sf A_{figura}=\dfrac{47}{3}\\\huge\boxed{\boxed{\boxed{\boxed{\sf A_{figura}=15 , 66~m^2\checkmark}}}}$