Question

calcule:
a imagem referente a questão está logo acima ajuda aí mano!​

Answer 1

Resposta:

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{58}{15}$

Explicação passo-a-passo:

Considerando $x^2+1=u$, temos que $\frac{du}{dx}=2x\therefore dx=\frac{1}{2x}\;du$. Como vamos trocar de variável, devemos recalcular os limites de integração. Para $x=0$, $u=0^2+1=1$ e, para $x=\sqrt{3}$, $u=(\sqrt{3})^2+1=4$. Aplicando:

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\int_1^{4}x^3\sqrt{u}\cdot \frac{1}{2x}\;du$

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{1}{2}\int_1^{4}x^2\sqrt{u}\;du$

Sendo $x^2=u-1$:

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{1}{2}\int_1^{4}(u-1)\sqrt{u}\;du$

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{1}{2}\int_1^{4}u\sqrt{u}-\sqrt{u}\;du$

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{1}{2}\int_1^{4}u^{3/2}-u^{1/2}\;du$

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{1}{2}\left[\frac{u^{3/2+1}}{3/2+1}-\frac{u^{1/2+1}}{1/2+1}\right]_1^4$

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{1}{2}\left[\frac{u^{5/2}}{5/2}-\frac{u^{3/2}}{3/2}\right]_1^4$

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{1}{2}\left[\frac{2}{5}\sqrt{4^5}-\frac{2}{3}\sqrt{4^3}-\left(\frac{2}{5}\sqrt{1^5}-\frac{2}{3}\sqrt{1^3}\right)\right]$

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{1}{2}\left[\frac{2}{5}\cdot32-\frac{2}{3}\cdot 8-\left(\frac{2}{5}-\frac{2}{3}\right)\right]$

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{1}{2}\cdot \frac{116}{15}$

$\int_0^{\sqrt{3}}x^3\sqrt{x^2+1}\;dx=\frac{58}{15}$

Answer 2

Explicação passo-a-passo:

$\int\limits^3_0 {x^{3}\sqrt{x^{2}+1}} \, dx$

Obs.: considere o limite superior 3 como $\sqrt{3}$, pois a fórmula do Brainly não está aceitando o radical $\sqrt{}$.

Solução:

- aplicar integração por substituição

      $u=x^{2}+1$   ;   $x^{2}=u-1$   ;   $\frac{du}{dx}=2x$  →  $dx=\frac{1}{2x}du$

- substituir

      $\int\limits^3_0 {x^{3}.\sqrt{u}.\frac{1}{2x}} \, du$

- simplificar x³ com x

      $\int\limits^3_0 {x^{2}.\sqrt{u}.\frac{1}{2}} \, du$

- remover a constante  $\frac{1}{2}$  e substituir x² por u - 1

      $\frac{1}{2}\int\limits^3_0 {(u-1).\sqrt{u}} \, du$

- multiplicar pela distributiva

      $\frac{1}{2}\int\limits^3_0 {u\sqrt{u}-\sqrt{u}} \, du$

      $\frac{1}{2}\int\limits^3_0 {\sqrt{u^{3}}-\sqrt{u}} \, du$

      $\frac{1}{2}\int\limits^3_0 {u^{\frac{3}{2}}-u^{\frac{1}{2}}} \, du$

- aplicar a regra da soma

      $\frac{1}{2}(\int\limits^3_0 {u^{\frac{3}{2}}} \, du-\int\limits^3_0 {u^{\frac{1}{2}}} \, du)$

- resolver a integral

      $\frac{1}{2}[\frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}-\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}]\left {\sqrt{3}} \atop {0}} \right.$

      $\frac{1}{2}[\frac{u^{\frac{5}{2}}}{\frac{5}{2}}-\frac{u^{\frac{3}{2}}}{\frac{3}{2}}]\left {{\sqrt{3}} \atop {0}} \right.$

      $\frac{1}{2}[\frac{2}{5}u^{\frac{5}{2}}-\frac{2}{3}u^{\frac{3}{2}}]\left {{\sqrt{3}} \atop {0}} \right.$

      $\frac{1}{2}[\frac{2}{5}\sqrt{u^{5}}-\frac{2}{3}\sqrt{u^{3}}]\left {{\sqrt{3}} \atop {0}} \right.$

- substituir u por x² + 1

      $\frac{1}{2}[\frac{2}{5}\sqrt{(x^{2}+1)^{5}}-\frac{2}{3}\sqrt{(x^{2}+1)^{3}}]\left {{\sqrt{3}} \atop {0}} \right.$

- calcular a integral definida

      $\frac{1}{2}[(\frac{2}{5}\sqrt{((\sqrt{3})^{2}+1)^{5}}-\frac{2}{3}\sqrt{((\sqrt{3})^{2}+1)^{3}})-(\frac{2}{5}\sqrt{(0^{2}+1)^{5}}-\frac{2}{3}\sqrt{(0^{2}+1)^{3}})]$

      $\frac{1}{2}[(\frac{2}{5}\sqrt{(3+1)^{5}}-\frac{2}{3}\sqrt{(3+1)^{3}})-(\frac{2}{5}\sqrt{1^{5}}-\frac{2}{3}\sqrt{1^{3}})]$

      $\frac{1}{2}[(\frac{2}{5}\sqrt{1024}-\frac{2}{3}\sqrt{64})-(\frac{2}{5}.1-\frac{2}{3}.1)]$

      $\frac{1}{2}[(\frac{2}{5}.32-\frac{2}{3}.8)-(\frac{2}{5}-\frac{2}{3})]$

      $\frac{1}{2}[(\frac{64}{5}-\frac{16}{3})-(-\frac{4}{15})]$

      $\frac{1}{2}[\frac{112}{15}+\frac{4}{15}]$

      $\frac{1}{2}.\frac{116}{15}=\frac{116}{30}=\frac{58}{15}$

Resposta:  $\frac{58}{15}$  ;  $em$ $decimal:$ $3,86666...$