Question

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questão 3

Questões de ensino superior.
cálculo diferencial e integral.

Answer 1

$\int\limits^3_{-1} {(2x^3-2x+3)} \, dx = \left [2* \frac{x^{3+1}}{3+1}-2* \frac{x^{1+1}}{1+1}+3x \right ]^3_{-1}\\\\ = \left [ \frac{x^4}{2}-x^2+3x \right]^3_{-1} \\\\= \left( \frac{3^4}{2}-3^2+3*(3)\right)- \left( \frac{(-1)^4}{2}-(-1)^2+3(-1)\right)=44$

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b)


$\int\limits^2_{-3} { \frac{y^3}{5} -y^2} \, dy = \int\limits^2_{-3} { \frac{y^3-5y^2}{5} } \, dy \\\\= \frac{1}{5} \int\limits^2_{-3} y^3-5y^2 \, dy = \frac{1}{5}\left[ \frac{y^4}{4}- \frac{5y^3}{3} \right]^2_{-3} = \frac{1}{5}\left[ \frac{3y^4-20y^3}{12} \right] ^2_{-3}\\\\=\frac{1}{5}\left[ \left( \frac{3*(2)^4-20*(2)^3}{12}\right) - \left(\frac{3*(-3)^4-20*(-3)^3}{12}\right) \right]= - \frac{179}{12}$