Question

6) Calcule as integrais:g)∫y³(2y²-3)dy;k)(∛x+1/∛x)dx;j)∫(2/x³+3/x²+5)dx

Answer 1

$g)\ \int y^3(2y^2-3).dy\\\\ =\int (2y^5-3y^3).dy\\\\ =2\dfrac{y^{5+1}}{5+1}-3\dfrac{y^{3+1}}{3+1}+constante\\\\ =2\dfrac{y^6}{6}-3\dfrac{y^{4}}{4}+constante\\\\ \boxed{=\dfrac{y^6}{3}-\dfrac{3y^{4}}{4}+constante\\\\}$



$k)\ {\displaystyle\int \left(\dfrac{\sqrt[3]{x}+1}{\sqrt[3]{x}}\right). dx\\\\ = {\displaystyle \int \left(\dfrac{\sqrt[3]{x}}{\sqrt[3]{x}}+\dfrac{1}{\sqrt[3]{x}}\right). dx\\\\ = {\displaystyle\int \left(1+\sqrt[3]{x^{-1}}\right). dx\\\\ =\int \left(1+x^{-\frac{1}{3}}\right). dx\\\\ =\dfrac{1x^{0+1}}{0+1}+\dfrac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+constante\\\\ =x+\dfrac{x^{\frac{2}{3}}}{\frac{2}{3}}+constante\\\\ =x+\dfrac{3x^{\frac{2}{3}}}{2}+constante\\\\ \boxed{=x+\dfrac{3\sqrt[3]{x^2}}{2}+constante}$


$j)\ {\displaystyle \int \left(\dfrac{2}{x^3}+\dfrac{3}{x^2}+5\right).dx\\\\ = {\displaystyle \int \left(2x^{-3}+3x^{-2}+5\right).dx\\\\ =\dfrac{2x^{-3+1}}{-3+1}+\dfrac{3x^{-2+1}}{-2+1}+\dfrac{5x^{0+1}}{0+1}+constante\\\\ =\dfrac{2x^{-2}}{-2}+\dfrac{3x^{-1}}{-1}+5x+constante\\\\ =-x^{-2}-3x^{-1}+5x+constante\\\\ =-\dfrac{1}{x^{2}}-\dfrac{1}{3x^{1}}+5x+constante\\\\ \boxed{=\dfrac{-1-3x+5x^3}{x^2}+constante\\\\}$


Espero ter ajudado.
Bons estudos!

Answer 2

Resposta:

ScreenBlackAmbicioso

g)\ \int y^3(2y^2-3).dy\\\\
=\int (2y^5-3y^3).dy\\\\
=2\dfrac{y^{5+1}}{5+1}-3\dfrac{y^{3+1}}{3+1}+constante\\\\
=2\dfrac{y^6}{6}-3\dfrac{y^{4}}{4}+constante\\\\
\boxed{=\dfrac{y^6}{3}-\dfrac{3y^{4}}{4}+constante\\\\}

k)\ ${\displaystyle\int \left(\dfrac{\sqrt[3]{x}+1}{\sqrt[3]{x}}\right). dx\\\\ =${\displaystyle \int \left(\dfrac{\sqrt[3]{x}}{\sqrt[3]{x}}+\dfrac{1}{\sqrt[3]{x}}\right). dx\\\\ =${\displaystyle\int \left(1+\sqrt[3]{x^{-1}}\right). dx\\\\ =\int \left(1+x^{-\frac{1}{3}}\right). dx\\\\ =\dfrac{1x^{0+1}}{0+1}+\dfrac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+constante\\\\ =x+\dfrac{x^{\frac{2}{3}}}{\frac{2}{3}}+constante\\\\ =x+\dfrac{3x^{\frac{2}{3}}}{2}+constante\\\\ \boxed{=x+\dfrac{3\sqrt[3]{x^2}}{2}+constante}

j)\ ${\displaystyle \int \left(\dfrac{2}{x^3}+\dfrac{3}{x^2}+5\right).dx\\\\
=${\displaystyle \int \left(2x^{-3}+3x^{-2}+5\right).dx\\\\
=\dfrac{2x^{-3+1}}{-3+1}+\dfrac{3x^{-2+1}}{-2+1}+\dfrac{5x^{0+1}}{0+1}+constante\\\\
=\dfrac{2x^{-2}}{-2}+\dfrac{3x^{-1}}{-1}+5x+constante\\\\
=-x^{-2}-3x^{-1}+5x+constante\\\\
=-\dfrac{1}{x^{2}}-\dfrac{1}{3x^{1}}+5x+constante\\\\
\boxed{=\dfrac{-1-3x+5x^3}{x^2}+constante\\\\}

Espero ter ajudado.

Bons estudos!

Explicação passo-a-passo: