Question

encontre a área da região limitada pela curva y=2x+1 e y=x²

Answer 1

Vamos começar determinando os pontos de intersecção das duas curvas:

$2x+1~=~x^2\\\\\\x^2-2x-1~=~0\\\\\\\Delta~=~(-2)^2-4.1.(-1)~=~4+4~=~\boxed{8}\\\\\\\boxed{x'~=~1+\sqrt{2}}\\\\\\\boxed{x''~=~1-\sqrt{2}}$

Neste intervalo (x'' ≤ x ≤ x') a curva y = 2x+1  está acima da curva y = x²  (ver anexo).

Sendo assim, podemos calcular a área desta região por:

$Area~=~\int\limits_{1-\sqrt{2}}^{1+\sqrt{2}}\left(2x+1\right)dx~-~\int\limits_{1-\sqrt{2}}^{1+\sqrt{2}}\left(x^2\right)dx\\\\\\Area~=~\int\limits_{1-\sqrt{2}}^{1+\sqrt{2}}\left(2x+1-x^2\right)dx\\\\\\Area~=~\left(\frac{2x^2}{2}+\frac{1x^1}{1}-\frac{x^3}{3}\right|_{1-\sqrt{2}}^{1+\sqrt{2}}\\\\\\Area~=~\left((1+\sqrt{2})^2~+~(1+\sqrt{2})~-~\frac{(1+\sqrt{2})^3}{3}\right)~-~\left((1-\sqrt{2})^2~+~(1-\sqrt{2})~-~\frac{(1-\sqrt{2})^3}{3}\right)$

$Area~=~\left(3+2\sqrt{2}+1+\sqrt{2}~-~\frac{7+5\sqrt{2}}{3}\right)~-~\left(3-2\sqrt{2}+1-\sqrt{2}~-~\frac{7-5\sqrt{2}}{3}\right)\\\\\\Area~=~6\sqrt{2}-\frac{10\sqrt{2}}{3}\\\\\\\boxed{Area~=~\frac{8\sqrt{2}}{3}~unidades~de~area}$