Question

Suponha que para resolver um problema, seja necessário primeiro calcular a área entre as curvas y = 2 e y = x2-2. Sabendo que x varia de [-2, 2], determine a área entre essas curvas e assinale a alternativa que corresponde a área aproximada. A . 20,20 u.a. B. 10,67 u.a. C. 8,40 u.a. D. 14,33 u.a E. 5.33 u.a​

Answer 1

$\large\boxed{\begin{array}{l}\underline{\rm Observe\,a\,figura\,que\,eu\,anexei.}\\\sf veja\,que\,calcular\,a\,\acute area\,entre\,as\,curvas\,y=2\,e\,y=x^2-2\\\sf no\,intervalo\,-2\,a\,2\,\acute e\,o\,mesmo\,que\,calcular\\\sf a\,\acute area\,entre\,as\,curvas\,no\,intervalo\,0\,e\,2\,e\,multiplicar\,por\,2.\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf A=2\int_0^2(2-[x^2-2])dx\\\displaystyle\sf A=2\int_0^2(2-x^2+2)dx\\\displaystyle\sf A=2\int_0^2(4-x^2)\,dx\\\\\sf A=\bigg[8x-\dfrac{2}{3}x^3\bigg]_0^2\\\sf n\tilde ao\,precisa\,substituir\,x\,por\,0\,pois\,tudo\,se\,anula.\\\sf A=8\cdot2-\dfrac{2}{3}\cdot2^3\\\\\sf A=16-\dfrac{16}{3}\\\\\sf A=\dfrac{48-16}{3}\\\\\sf A=\dfrac{32}{3}\,u\bullet a\\\\\sf A=10,67\,u\bullet a \end{array}}$