Question

resolva a integral indefinida ​

Answer 1

Resposta:

Explicação passo-a-passo:

$\int\limits^2_0 {\sqrt{x^3} \, dx = \int\limits^2_0x^{\frac{3}{2}}dx=\\\\\frac{x^{\frac{5}{2}}}{\frac{5}{2}}} = \frac{2*\sqrt{x^5}}{5} \\\\Substituindo \ os \ limites \\\\\\ \frac{2*\sqrt{2^5}}{5} - \frac{2\sqrt{0^5}}{5} = \frac{2*\sqrt{32}}{5}-0 = \frac{2*4\sqrt{2}}{5} = \frac{8}{5}\sqrt{2}$

Answer 2

• O resultado da sua integral definida é 8√2/5.

Em uma integral definida devemos primeiramente resolver a integral e depois aplicar o ( TFC ) ⇔ Teorema Fundamental do Cálculo. Logo:

$\large\displaystyle\text{ \begin{aligned}\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \int_0^2 x^{\frac{3}{2}} \ dx\end{aligned} }$

Perceba que caimos em um caso de integração imediato, que é a integração de um monômio. Dada por:

$\large\displaystyle\text{ \begin{aligned}\int x^n\ dx \Leftrightarrow \frac{x^{n+1}}{n+1}+k\ , \ \underline{\underline{n\neq -1}}\end{aligned} }$

• Logo, aplicando a integral do monômio, temos que:

$\large\displaystyle\text{ \begin{aligned}\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \int_0^2 x^{\frac{3}{2}} \ dx\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned}\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \int_0^2 \frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} \ dx\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned}\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \int_0^2 \frac{x^{\frac{5}{2}}}{\frac{5}{2}} \ dx\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned}\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \int_0^2 \sqrt{x^{5}} \cdot \frac{2}{5}\ dx\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned}\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \int_0^2 \frac{2\sqrt{x^{5}}}{5}\ dx\end{aligned} }$

• Agora, devemos aplicar o Teorema Fundamental do Cálculo.

$\large\displaystyle\text{ \begin{aligned}\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \left.\frac{2\sqrt{x^{5}}}{5}\right|^2_0\ \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned}\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \frac{2\sqrt{(2)^{5}}}{5} - \frac{2\sqrt{(0)^{5}}}{5}\ \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned}\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \frac{2\sqrt{32}}{5} - 0\ \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned}\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \frac{2\cdot4\sqrt{2}}{5} - 0\ \end{aligned} }$

$\therefore \large\displaystyle\text{ \begin{aligned}\boxed{\boxed{\green{\int_0^2 \sqrt{x^3} \ dx\Leftrightarrow \frac{8\sqrt{2}}{5}}}} \ \end{aligned} }$

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Integrais definidas.

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