Question

Por favor me ajude nesta questão

672b6f36e1522c5e65efede0f3449920.docx

Answer 1

A questão pede para calcular a área entre as curvas

$y=\dfrac{x^{2}}{2}\;\;\text{ e }\;\;y^{2}=2x$

no intervalo $0\leq x \leq 2.$


Reescrevendo as curvas como funções de $x,$ no intervalo considerado:

$f(x)=\dfrac{x^{2}}{2}\;\;\text{ e }\;\;g(x)=\sqrt{2x}$


Para $x \in [0;\,2],$ temos que

$g(x)\geq f(x)\\ \\ \sqrt{2x} \geq \dfrac{x^{2}}{2}$


Logo, área da região hachurada é

$A=\int_{a}^{b}{[g(x)-f(x)]\,dx}\\ \\ \\ A=\int_{0}^{2}{\left[\sqrt{2x}-\dfrac{x^{2}}{2} \right]\,dx}\\ \\ \\ A=\int_{0}^{2}{\sqrt{2}\cdot\sqrt{x}\,dx}-\int_{0}^{2}{\dfrac{x^{2}}{2}\,dx}\\ \\ \\ A=\sqrt{2}\int_{0}^{2}{x^{1/2}\,dx}-\dfrac{1}{2}\int_{0}^{2}{x^{2}\,dx}\\ \\ \\ A=\sqrt{2}\left[\dfrac{x^{(1/2)+1}}{\frac{1}{2}+1}\right]_{0}^{2}-\dfrac{1}{2}\left[\dfrac{x^{2+1}}{2+1} \right ]_{0}^{2}\\ \\ \\ A=\sqrt{2}\left[\dfrac{x^{3/2}}{(\frac{3}{2})}\right]_{0}^{2}-\dfrac{1}{2}\left[\dfrac{x^{3}}{3} \right ]_{0}^{2}\\ \\ \\ A=\sqrt{2}\cdot \dfrac{2}{3}\left[x^{3/2}\right]_{0}^{2}-\dfrac{1}{2}\cdot \dfrac{1}{3}\left[x^{3} \right ]_{0}^{2}$


$A=\dfrac{2\sqrt{2}}{3}\left[x^{3/2}\right]_{0}^{2}-\dfrac{1}{6}\left[x^{3} \right ]_{0}^{2}\\ \\ \\ A=\dfrac{2\sqrt{2}}{3}\left[\sqrt{x^{3}}\right]_{0}^{2}-\dfrac{1}{6}\left[x^{3} \right ]_{0}^{2}\\ \\ \\ A=\dfrac{2\sqrt{2}}{3}\left[\sqrt{2^{3}}-\sqrt{0^{3}}\right]-\dfrac{1} {6}\left[2^{3}-0^{3} \right ]\\ \\ \\ A=\dfrac{2\sqrt{2}}{3}\left[\sqrt{2^{2}\cdot 2}-0\right]-\dfrac{1}{6}\left[8-0 \right ]\\ \\ \\ A=\dfrac{2\sqrt{2}}{3}\left[\sqrt{2^{2}}\cdot \sqrt{2}\right]-\dfrac{1}{6}\left[8 \right ]\\ \\ \\ A=\dfrac{2\sqrt{2}}{3}\left[2\sqrt{2}\right]-\dfrac{8}{6}\\ \\ \\ A=\dfrac{8}{3}-\dfrac{4}{3}\\ \\ \\ A=\dfrac{4}{3}\text{ u.a.}$


Resposta: 2ª alternativa: $4/3.$

Answer 2

$Ola'~~~\mathbb{IRISLANE} \\ \\ Temos~duas~fun\zeta\tilde{o}es: \\ \\\boxed{ y= \frac{ x^{2} }{2}} \\ \\ y^{2} =2x ~~--\ \textgreater \ deste~isolamos~(y)~seria: \\ \boxed{y= \sqrt{2x} } \\~intervalo~\boxed{0~\le~x~\le~ 2}  \\ \\ \mathbb{iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii} \\ Para~calcular~a~\acute{a}rea~usaremos~a~integral~veja:$


$\int_{0} ^{2} ( \sqrt{2x} - \frac{ x^{2} }{2} )dx~~---\ \textgreater \ desenvolvendo: \\ \\ \int_{0} ^{2} \sqrt{2x} dx-\int_{0} ^{2} \frac{ x^{2} }{2} dx \\ \\ \sqrt{2}\int_{0} ^{2} \sqrt{x} dx- \frac{1}{2} \int_{0} ^{2} x^{2} dx ~~--\ \textgreater \ integrando~temos: \\ \\ \sqrt{2}. \frac{ x^{ 3/2 } }{3/2} - \frac{1}{2} \frac{ x^{3} }{3} ~\bigg] _{0} ^{2}$

$\frac{ \sqrt{2.x ^{3} } }{3/2} - \frac{ x^{3} }{6} ~\bigg]_{0} ^{2} \\ \\ \frac{2 \sqrt{2 x^{3} } }{3}- \frac{ x^{3} }{6}~\bigg]_{0} ^{2}~~---\ \textgreater \ substituindo~os~valores~temos: \\ \\ \frac{2 \sqrt{2. 2^{3} } }{3} - \frac{ 2^{3} }{6} +~0 \\ \\ \frac{8}{3} - \frac{8}{6} \\ \\ \boxed{\boxed{\frac{4}{3}u ^{2}}} \\ \\\mathbb{iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii} \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~ter~ajudado!!$