Question

Veja na foto a Pergunta.
É de Matemática.

Answer 1

Resposta:

Essa vai dar trabalho. vou fazer no caderno, tiro uma foto e posto, tá bom assim. vai demorar um pouco. daqui a uns 40 minutos.

Explicação passo a passo:

muito trabalho, fiz bem caprichado.

acho que vai dar para entender.

veja a foto.

Answer 2

Observe a solução de cada item:

$\begin{aligned}& a)\qquad\,\int^3_{-1}(3x+5x^2+2x^5)dx~~\Longleftrightarrow\\\\& \Longleftrightarrow~~\int3xdx+\int5x^2dx+\int2x^5dx~~\Longleftrightarrow\\\\& \Longleftrightarrow~~\dfrac{3x^2}{2}+\dfrac{5x^3}{3}+\dfrac{2x^6}{6}+C~~\Longleftrightarrow\\\\& \Longleftrightarrow~~\dfrac{3x^2}{2}+\dfrac{5x^3}{3}+\dfrac{x^6}{3}+C.\end{aligned}$

Pelo TFC (Teorema Fundamental do Cálculo) :

$\begin{aligned}&\int^3_{-1}(3x+5x^2+2x^5)dx=\dfrac{3x^2}{2}+\dfrac{5x^3}{3}+\dfrac{x^6}{3}+C\,\Bigg|^3_{-1}\end{aligned}$

                         $\begin{aligned}&\qquad\qquad\qquad\qquad\qquad=\dfrac{3(3)^2}{2}+\dfrac{5(3)^3}{3}+\dfrac{(3)^6}{3}+C-\bigg(\dfrac{3(-1)^2}{2}+\dfrac{5(-1)^3}{3}+\dfrac{(-1)^6}{3}+C\bigg)\\\\&\qquad\qquad\qquad\qquad\qquad=\dfrac{27}{2}+45+3^5+C-\dfrac{3}{2}+\dfrac{5}{3}-\dfrac{1}{3}-C\\\\&\qquad\qquad\qquad\qquad\qquad=13,5+45+243+0-1,5+\dfrac{4}{3}\\\\&\qquad\qquad\qquad\qquad\qquad=300+\dfrac{4}{3}\\\\&\qquad\qquad\qquad\qquad\qquad=\dfrac{900+4}{3}\\\\&\qquad\qquad\qquad\qquad\qquad=\boxed{\dfrac{904}{3}}\end{aligned}$

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$\begin{aligned}& b)\qquad\,\int\bigg(e^x+\dfrac{1}{x}\bigg)dx~~\Longleftrightarrow\\\\& \Longleftrightarrow~~\int e^xdx+\int\dfrac{1}{x}dx~~\Longleftrightarrow\\\\& \Longleftrightarrow~~\boxed{e^x+ln|x|+C.}\end{aligned}$

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$\begin{aligned}& c)\qquad\,\int\bigg(\dfrac{1}{x^2}+\dfrac{2}{x^3}+\sqrt[3]{x}\bigg)dx~~\Longleftrightarrow\\\\& \Longleftrightarrow~~\int\dfrac{1}{x^2}dx+\int\dfrac{2}{x^3}dx+\int\sqrt[3]{x}dx~~\Longleftrightarrow\\\\& \Longleftrightarrow~~-\dfrac{1}{x}-\dfrac{1}{x^2}+\int x^{\frac{1}{3}}dx~~\Longleftrightarrow\\\\& \Longleftrightarrow~~\boxed{-\dfrac{1}{x}-\dfrac{1}{x^2}+\dfrac{3x\sqrt[3]{x}}{4}+C.}\end{aligned}$

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$\begin{aligned}&d)\qquad\,\int sen(5x+4)dx\\\\& \implies~~u=5x+4\implies du/dx=5\Longleftrightarrow dx=du/5.\\\\&~~\qquad\,\int \dfrac{sen(u)du}{5}~~\Longleftrightarrow\\\\&\Longleftrightarrow~~\dfrac{1}{5}\int sen(u)du~~\Longleftrightarrow\\\\&\Longleftrightarrow~~-\dfrac{1}{5}cos(u)+C~~\Longleftrightarrow\\\\&\Longleftrightarrow~~\boxed{-\dfrac{1}{5}cos(5x+4)+C.}\end{aligned}$

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$\begin{aligned}&e)\qquad\,\int \dfrac{1}{7x+2}dx\\\\& \implies~~u=7x+2\implies du/dx=7\Longleftrightarrow dx=du/7.\\\\&~~\qquad\,\int \dfrac{du}{7u}~~\Longleftrightarrow\\\\&\Longleftrightarrow~~\dfrac{1}{7}\int \dfrac{du}{u}~~\Longleftrightarrow\\\\&\Longleftrightarrow~~\dfrac{1}{7}ln|u|+C~~\Longleftrightarrow\\\\&\Longleftrightarrow~~\boxed{\dfrac{1}{7}ln|7x+2|+C.}\end{aligned}$