Respostas

2014-05-23T00:38:38-03:00
 \int\limits^3_{-1} {3x+5x^2+2x^5} \, dx =[ \frac{3x^2}{2} +  \frac{5x^3}{3} + \frac{2x^6}{6} ]_{-1}^{3}\\\\\boxed{F(3) - F(-1)}\\\\F(3)=\frac{3*3^2}{2} +  \frac{5*3^3}{3} + \frac{2*3^6}{6}= \frac{603}{2} \\\\\\F(-1)=\frac{3(-1)^2}{2} +  \frac{5(-1)^3}{3} + \frac{2(-1)^6}{6}= \frac{1}{6} \\\\ resposta: \boxed{\frac{603}{2} - \frac{1}{6} = \frac{904}{3} }
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 \int\limits {e^x +\frac{1}{x} } \, dx
podemos reescrever como
 \int\limits {e^x +x^{-1} } \, dx =\boxed{e^x+ln|x| + K}

integral de x^{-1} = ln|x|
K = constante
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 \int\limits { \frac{1}{x^2} + \frac{2}{x^3}+  \sqrt[3]{x}    } \, dx
reescrevendo a fução temos
\int\limits { x^{-2}+2x^{-3}+x^{ \frac{1}{3}} \, dx=\boxed{ \frac{x^{-1}}{-1} + \frac{2x^{-2}}{-2} + \frac{x^{ \frac{4}{3}}}{ \frac{4}{3} } }= \boxed{ -\frac{1}{x} - \frac{1}{x^2} + \frac{3x^{ \frac{4}{3} }}{4} }

RESPOSTA: \boxed{\boxed{ -\frac{1}{x} - \frac{1}{x^2}+ \sqrt[3]{x^4} + K}}

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 \int\limits {sen(5x+4)} \, dx

u = 5x+4
du = 5 *dx (derivada de 5x+4=5)
du=5*dx\\\\ \frac{du}{5} =dx
então temos
 \int\limits{sen(5x+4)} \, dx
sabemos que 5x+4 = u..então susbtituimos o valor antes d integrar
e que dx = du/5
 \int\limits {sen(u)} .\frac{du}{5} = \frac{1}{5}  \int\limits {sen(u)} .du= \frac{1}{5} [-cos(u)]= \boxed{\frac{-cos(u)}{5} }

como u = 5x+4
RESPOSTA: \boxed{ \frac{-cos(5x+4)}{5}+K }
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 \int\limits { \frac{1}{7x+2} } \, dx

u = 7x+2
du = 7
du = 7dx\\\\ \frac{du}{7} =dx

substituindo os valores
 \int\limits { \frac{1}{7x+2} } \, dx = \int\limits { \frac{1}{u} } .  \frac{du}{7}=\\\\\\ = \frac{1}{7}  \int\limits {u^{-1}} \, dx =\boxed{ \frac{1}{7} *ln|u|}\\\\\\\\resposta: \boxed{\boxed{ \frac{ln|7x+2|}{7} +K}}


2 4 2
2014-05-23T00:44:18-03:00

lembrando-se que o intervá-lo é entre (3 e -1), não consegui editar o (-1) no Latex.
Resolução da letra A: 

   3 \frac{ x^{2} }{2} + 5 \frac{ x^{3} }{3}  + 2 \frac{ x^{6} }{6}
   3 \frac{ x^{2} }{2}  + 5 \frac{ x^{3} }{3} +  \frac{ x^{6} }{6}
  \left[\begin{array}{ccc}3. \frac{3^{2} }{2} + 5 \frac{ 3^{3} }{3 } +  \frac{ 3^{6} }{6}   \end{array}\right]     -\left[\begin{array}{ccc}3.  \frac{ (-1)^{2} }{2} + 5. \frac{ (-1)^{3} }{3}  + \frac{ (-1)^{6} }{6} \end{array}\right]

  \left[\begin{array}{ccc} \frac{27}{2} +  \frac{135}{3} + \frac{729}{3}  \end{array}\right]    - \left[\begin{array}{ccc} \frac{3}{2} +  \frac{5}{3} +  \frac{1}{6}   \end{array}\right]
  \left[\begin{array}{ccc} \frac{27}{2} + 288 \end{array}\right]    \left[\begin{array}{ccc} \frac{+1}{6} \end{array}\right]

 \frac{603}{2} - \frac{1}{6} =  \frac{904}{3}

Letra B:
 e^{x} + ln(x) + C

Letra C:
 \int\limits { x^{-2} } \, dx  + 2  \int\limits  x^{-3} \ dx +  \int\limits  x^{ \frac{1}{3} }   \, dx
 \int\limits   \frac{x^{-2+1}}{-2 +1}    \, dx   \int\limits 2 \frac{ x^{-3+1} }{-3+1}  \, dx +  \int\limits  x^{  \frac{\frac{1}{3}+1}{ \frac{1}{3}+1 }  }  \, dx
  \left[\begin{array}{ccc} \frac{-1}{x} + 2. \frac{ x^{-2} }{-2}+  \frac{3}{4}  x^{ \frac{3}{4} }  \end{array}\right]

  \left[\begin{array}{ccc} \frac{-1}{x} - {x^{-2}+ \frac{3}{4}  \sqrt[3]{x^{4} }  \end{array}\right]
  \left[\begin{array}{ccc} \frac{-1}{x} - \frac{1}{ x^{2}}+ \frac{3}{4} \sqrt[3]{ x^{4} }    \end{array}\right]

Letra D:
Há uma regrinha que eu sei:A integral de Sen (ax)= - cos (ax)/a
- cos \frac{(5x + 4) }{5} + c