Question

Como resolver passo a passo a integral a seguir: de 3 a -3 S x^2-sx^3
=16dx

Answer 1

Olá

Primeiro vamos "quebrar" a integral em 3 integrais

$\displaystyle \int\limits^3_ {-3}x^2-2x^3+16 \, dx \\ \\ \\ \int\limits^3_ {-3}x^2\, dx ~~~~ ~-\int\limits^3_ {-3}2x^3 \, dx ~~~~~+\int\limits^3_ {-3}16 \, dx \\ \\ \\ \text{Tira as constantes de dentro das integrais para facilitar} \\ \\ \\ \int\limits^3_ {-3}x^2\, dx ~~~~ ~-2\int\limits^3_ {-3}x^3 \, dx ~~~~~+16\int\limits^3_ {-3} \, dx \\ \\ \\ \text{Agora e so resolve a integral pela regra da tabela} \boxed{\int\limits {x}^p \, dx = \frac{x^p^+^1}{p+1} }$

$\displaystyle = \frac{x^2^+^1}{2+1} ~-~2\cdot( \frac{x^3^+^1}{3+1} )~+~16\cdot( \frac{x^0^+^1}{0+1} ) \\ \\ \\ = \frac{x^3}{3}~-~2 \cdot\frac{x^4}{4} ~+~16x \\ \\ \\ \text{simplifica} \\ \\ = \frac{x^3}{3}~-~\diagup\!\!\!\!2 \cdot\frac{x^4}{\diagup\!\!\!\!4} ~+~16x$

$\displaystyle =\left.\left(\dfrac{x^3}{3}-\dfrac{x^4}{2}+16x\,\mathrm{\,}\right)\right|_{-3}^{3} \\ \\ \\ \text{Agora e so substituir os valores, limite superior menos limite inferior}\\ \\ \\ =\left.\left(\dfrac{3^3}{3}-\dfrac{3^4}{2}+16\cdot(3)\,\mathrm{\,}\right)~~-~~\left.\left(\dfrac{(-3)^3}{3}-\dfrac{(-3)^4}{2}+16\cdot(-3)\,\mathrm{\,}\right)$
$\displaystyle= \left.\left( \frac{33}{2} \mathrm{\,}\right) -\left.\left( -\frac{195}{2} \mathrm{\,}\right)$
$\displaystyle =\frac{33}{2} + \frac{195}{2} \\ \\ =\frac{33+195}{2} \\ \\ =\frac{228}{2} \\ \\ \boxed{\boxed{=114}}$