Question

Calcule a área da região compreendida entre:
(a) (1,5 pontos) f(x) = 2√x e g(x) = x.

(b) f(x) = x2 + 3 e g(x) = 5 − x2.

Answer 1

$\boxed{\begin{array}{l}\rm pontos~de~intersecc_{\!\!,}\tilde ao:\\\rm x=2\sqrt{x}\\\rm x^2=4x\\\rm x^2-4x=0\\\rm x(x-4)=0\\\rm x=0\\\rm x-4=0\\\rm x=4\\\displaystyle\rm A=\int_0^4(2\sqrt{x}-x)dx \\\\\displaystyle\rm A=\bigg[\dfrac{4}{3}x^{\frac{3}{2}}-\dfrac{1}{2}x^2\bigg]_0^4 \\ \\\rm A=\dfrac{4}{3}\cdot 4^{\frac{3}{2}}-\dfrac{1}{2}4^2\\\\\rm A=\dfrac{4}{3}\cdot(2^2)^{\frac{3}{2}}-\dfrac{1}{2}\cdot16\\\\\rm A=\dfrac{32}{3}-\dfrac{16}{2}\\\\\rm A=\dfrac{64-48}{6}\\\\\rm A=\dfrac{16\div2}{6\div2}\\\\\rm A=\dfrac{8}{3}~u\bullet a\end{array}}$

$\boxed{\begin{array}{l}\rm pontos~de~intersecc_{\!\!,}\tilde ao:\\\rm x^2+3=5-x^2\\\rm x^2+x^2=5-3\\\rm 2x^2=2\div2\\\rm x^2=1\\\rm x=\pm\sqrt{1}\\\rm x=\pm1\\\rm x= \implies y=4\\\rm x=-1\implies y=4\\\displaystyle\rm A=2\int_0^1(5-x^2-[x^2+3])dx\\\\\displaystyle\rm A=2\int_0^1(5-x^2-x^2-3)dx\\\\\displaystyle\rm A=2\int_0^1(2-x^2)dx\\\\\rm A=\bigg[4x-\dfrac{2}{3}x^3\bigg]_0^1\\\\\rm A=4\cdot1-\dfrac{2}{3}\cdot1^3\\\\\rm A=4-\dfrac{2}{3}\\\\\rm A=\dfrac{12-2}{3}\\\\\rm A=\dfrac{10}{3}~u\bullet a\end{array}}$