Question

A área limitada superiormente pelo gráfico de y = f(x) e inferiormente por y = g(x) é dada pela integra

Calcule a área sombreada na figura.

Answer 1

$\bmatrix g(x)=x^2-2\\g(x)=x \end$

primeiro encontrando o intervalo "a" "b" igualando as funçoes
$g(x)=f(x)\\\\x^2-2=x\\\\\boxed{x^2-x-2=0}$

resolvendo essa equação do segundo grau vc encontra as raízes
x'=-1 , x''=2
então
a= -1 , b=2
::::::::::::::::::::::::::::::::::::::::::::::::::::::::
$A= \int\limits^b_a {f(x)-g(x)} \, dx \\\\A= \int\limits^2_{-1} {(x)-(x^2-2)} \, dx \\\\ A= \int\limits^2_{-1} x-x^2+2 \; dx \\\\A= \left \frac{x^{1+1}}{1+1}- \frac{x^{2+1}}{2+1} +2x \right |^{2}_{-1}\\\\ A= \left \frac{x^{2}}{2}- \frac{x^{3}}{3} +2x \right |^{2}_{-1}\\\\ A= ( \frac{2^{2}}{2}- \frac{2^{3}}{3} +2*2) - ( \frac{(-1)^{2}}{2}- \frac{(-1)^{3}}{3} +2*(-1))\\\\\\ A= (2- \frac{8}{3}+4) - ( \frac{1}{2} - \frac{-1}{3}-2) \\\\A= \frac{10}{3} -( \frac{-7}{6} )$

$A= \frac{10}{3} + \frac{7}{6} = \frac{9}{2}$

Answer 2

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$\underline{\frak{ pontos~de~intersecc_{\!\!,}\tilde ao:}}\\\sf x^2-2=x\\\sf x^2-x-2=0\\\sf\Delta=1+8=9\\\sf x=\dfrac{1\pm3}{2}\begin{cases}\sf x_1=\dfrac{1+3}{2}=\dfrac{4}{2}=2\\\sf x_2=\dfrac{1-3}{2}=-\dfrac{2}{2}=-1\end{cases}\\\displaystyle\sf A=\int_{-1}^2( x-[x^2-2])dx\\\displaystyle\sf A=\int_{-1}^2(x+2-x^2)dx\\\sf A=\dfrac{1}{2}x^2+2x-\dfrac{1}{3}x^3\Bigg|_{-1}^2$

$\sf A=\dfrac{1}{2}\cdot 2^2+2\cdot 2-\dfrac{1}{3}\cdot 2^3-\left[\dfrac{1}{2}\cdot(-1)^2+2\cdot(-1)-\dfrac{1}{3}\cdot(-1)^3\right]\\\sf A=\dfrac{4}{2}+4-\dfrac{8}{3}-\dfrac{1}{2}+2-\dfrac{1}{3}\\\sf A=\dfrac{12+24-16-3+12-2}{6}\\\sf A=\dfrac{27\div3}{6\div3}\\\huge\boxed{\boxed{\boxed{\boxed{\sf A=\dfrac{9}{2}~~u\cdot a}}}}$