$\int$ Capítulo 5 - Seção 5.4 - Integrais Indefinidas e o Teorema da Variação Total (365)

Questão 19 - Jailson Bezerra
$\int (\cos {x} + \dfrac {1}{2} \cdot x) \cdot dx$

Solução
$\sin {x} + \dfrac {x^2}{4} + c$

Questões 21 – 46: Calcule a integral.

Questão 22- Robson Santos
$\displaystyle \int_{1}^{3} (1+2x-4x^3)dx$

Solução:
$\displaystyle \int_{1}^{3} (1+2x-4x^3)dx=$
$\displaystyle \int_{1}^{3} (x+x^2-x^4)dx=$
$\displaystyle (3+3^2-3^4)-(1+1^2-1^4)=(3+9-81)-(-1)=-70$



Questão 23 - Guilherme Fernandes
$\displaystyle \int_{0}^{-2} \left(\frac{1}{2}\,t^{4} + \frac{1}{4}\,t^{3} - t \right)$

Solução:
$\left[\dfrac{1}{2}\,\dfrac{t^{5}}{5} + \dfrac{1}{4}\,\dfrac{t^{4}}{4} - \dfrac{t^{2}}{2} \right]_{0}^{-2} = \left[\dfrac{t^{5}}{10} + \dfrac{t^{4}}{16} - \dfrac{t^{2}}{2} \right]_{0}^{-2} = \left[\dfrac{32t^{5} + 20t^{4} - 160t^{2}}{320} \right]_{0}^{-2} \\ 0 - \left(\dfrac{32\,(-2)^{5} + 20\,(-2)^{4} - 160\,(-2)^{2}}{320} \right) = 0 - \left(\dfrac{-1024 + 320 - 640}{320} \right) = \dfrac{21}{5}.$

Questão 24- Diana Keli
$\displaystyle\int_{0}^{3}(1+6w^{2}-10w^{4})dw$

Solução
$\displaystyle\int_{0}^{3}(1+6w^{2}-10w^{4})dw$
$\displaystyle\int_{0}^{3}\left ( w+\frac{6w^{3}}{3}-\frac{10w^{5}}{5} \right )dw$
$\displaystyle\int_{0}^{3}(w+2w^{3}-2w^{5})dw$

Substituindo w por 3 temos;
$(3+2\cdot27-2\cdot243)=$
$(3+54-486)= -429$

A substituição de w por 0 = 0, logo :
$-429-0=-429$
Questão 25- José Hudson
$\int_{0}^{2} (2x - 3)(4x^{2}+1)dx$

Solução : 
$\int_{0}^{2}(8^{3}+ 2x-12x^{2}-3)= \left[2x^{4}+ x^{2}-4x^{3}-3x \right]_{0}^{2}$
$(32+4-32-6)-(0) = -2$

Questão 28 - Guilherme Fernandes
$\displaystyle \int_{1}^{2}\left(\frac{1}{x^{2}} - \frac{4}{x^{3}} \right)\, dx$

Solução:
$\displaystyle \int_{1}^{2}(x^{-2} - 4x^{-3})\, dx = \left[\dfrac{x^{-1}}{-1} - \dfrac{4x^{-2}}{-2} \right]_{1}^{2} = \left[\dfrac{-2x^{-1} + 4x^{-2}}{2} \right]_{1}^{2} \\ \left(\dfrac{-2 \cdot 2^{-1} + 4 \cdot 2^{-2}}{2} \right) - \left(\dfrac{-2 \cdot 1^{-1} + 4 \cdot 1^{-2}}{2} \right) = 0 - \dfrac{2}{2} = -1.$
Questão 29 - José Hudson
$\int{1}^{4}\left(\dfrac{4+6u}{\sqrt{u}}\right)du$

Solução:
$\int_{1}^{4}(\dfrac{4}{\sqrt{u}}+\dfrac{6u}{\sqrt{u}})du$
$\int_{1}^{4}(4u^{\dfrac{-1}{2}}+6u\cdot u^{\dfrac{-1}{2}})du$
$\left[\dfrac{4u^{\dfrac{1}{2}}}{\dfrac{1}{2}}+\dfrac{6u^{\dfrac{3}{2}}}{\dfrac{3}{2}}\right]_{1}^{4}= \left[8u^{\dfrac{1}{2}}+ 4u^{\dfrac{3}{2}}\right]_{1}^{4}$
$(8\sqrt{4}+4\sqrt{4^{3}})-(8\sqrt{1}+4\sqrt{1^{3}})= 36$



Questão 33 - Antônio Wagner 
$\displaystyle\int_{1}^{2}(\frac{x}{2}-\frac{2}{x})dx$

Solução
$\displaystyle\int_{1}^{2}\frac{1}{4}x^{2}-2lnx$
$\displaystyle\int_{1}^{2}(1 - 2 ln 2) - (\frac{1}{4}-2 ln 1) = \frac{3}{4} - 2 ln 2$



Questão 35- Marden Torres

 $\displaystyle\int_{0}^{1}(x^{10} + 10^{x})dx$

Solução 

 $\displaystyle\int_{0}^{1}(x^{10} + 10^{x})dx = \Big[\frac{x^{11}}{11} + \frac{10^{x}}{ln 10}\Big]_{0}^{1}=$$\Big(\frac{1}{11} + \frac{10}{ln 10}\Big) -$$\Big(0 + \frac{1}{ln 10}\Big) = \frac{1}{11} + \frac{9}{ln 10}$
 


Questão 37- Diana Keli
$\displaystyle\int_{0}^{\pi/4}\frac{1+\cos^{2}\theta }{\cos^{2}\theta }d\theta$

Solução
$\displaystyle\int_{0}^{\pi/4}\frac{1+\cos^{2}\theta }{\cos^{2}\theta }d\theta$
$\displaystyle\int_{0}^{\pi/4}\left ( \frac{1}{\cos^{2}\theta }+\frac{cos^{2}\theta}{cos^{2}\theta} \right )d\theta$
$\displaystyle\int_{0}^{\pi/4}[\tan\theta+\theta]=$
$\displaystyle\left ( \tan\frac{\pi}{4}+\frac{\pi}{4} \right )-(0+0)=$
$\displaystyle1+\frac{\pi}{4}$


Questão 38 - Robson Santos
$\displaystyle\int_{0}^{\frac{\pi}{3}}\frac{sen\theta +sen\theta \cdot tg^2\theta}{sec^2\theta}d\theta$

Solução

$\displaystyle\int_{0}^{\frac{\pi}{3}}\frac{sen\theta +sen\theta \cdot tg^2\theta}{sec^2\theta }d\theta =$

$\displaystyle\int_{0}^{\frac{\pi}{3}}\frac{sen\theta (1+tg^2\theta )}{sec^2\theta }d\theta  =$

$\displaystyle\int_{0}^{\frac{\pi}{3}}\frac{sen\theta \cdot sec^2\theta  }{sec^2\theta }d\theta  =$

$\displaystyle\int_{0}^{\frac{\pi}{3}}sen\theta d\theta=$

$\displaystyle\int_{0}^{\frac{\pi}{3}}-cos\theta d\theta= \frac{-1}{2} -(-1)=\frac{1}{2}$

Questão 39 - Antônio Wagner 
$\displaystyle\int_{1}^{64}\frac{1+\sqrt[3]{x}}{\sqrt{x}}dx$

Solução
$\displaystyle\int_{1}^{64}\left(\frac{1}{x\frac{1}{2}}+\frac{x\frac{1}{3}}{x\frac{1}{2}}\right)dx$
$\displaystyle\int_{1}^{64}\left(x^\frac{-1}{2}+x^\frac{-1}{6}\right)dx = \int_{1}^{64}2x^\frac{1}{2}+\frac{6}{5}x^\frac{5}{6}$
$\displaystyle\int_{1}^{64} \left(16+\frac{192}{5}\right)-\left(2+\frac{6}{5}\right) = 14+\frac{186}{5} = \frac{256}{5}$

Questão 41- Robson Santos
$\displaystyle\int_{0}^{\frac{\sqrt{3}}{2}}\frac{dr}{\sqrt{1-r^2}}$

Solução
$\displaystyle\int_{0}^{\frac{\sqrt{3}}{2}}\frac{dr}{\sqrt{1-r^2}}=$
$\displaystyle\int_{0}^{\frac{\sqrt{3}}{2}}arcsen r = arcsen\left ( \frac{\sqrt{3}}{2} \right )- arcsen 0 = \frac{\pi}{3} - 0 = \frac{\pi}{3}$

Questão 42 - Jailson Bezerra
$\int_{1}^{2}  \dfrac {(x - 1)^3}{x^2} \cdot dx$

Solução
$\int_{1}^{2}  \dfrac {(x - 1)^3}{x^2} \cdot dx = \left [(x^3-3x^2+3x-1) \cdot x^{-2} \right ]_{1}^{2} = x - 3 + 3x^{-1}-x^{-2} = \dfrac {x}{2} -3x + 3 \cdot \ln {x} + \dfrac {1}{x} = \dfrac {2}{2} - 3 \cdot 2 + 3 \ln {2} + \dfrac {1}{2} - \left (\frac {1}{2} - 3 \cdot 1 + 3 \cdot \ln {1} + 1 \right ) = -2 + 3 \cdot \ln {2}$


Questão 43- Marden Torres
$\displaystyle\int_{0}^{1/\sqrt{3}}\frac{t^{2} - 1}{t^{4}-1}$

Solução

$\displaystyle\int_{0}^{1/\sqrt{3}}\frac{t^{2} - 1}{t^{4}-1}dt =$$\displaystyle\int_{0}^{1/\sqrt{3}}\frac{t^{2} - 1}{(t^{2} + 1)(t^{2} - 1)}dt =$

$\displaystyle\int_{0}^{1/\sqrt{3}}\frac{1}{t^{2} + 1}dt = \Big[\arctan t\Big]_{0}^{1/\sqrt{3}}=$$\arctan(1/\sqrt{3} - \arctan 0 = \displaystyle\frac{\pi}{6} - 0 = \frac{\pi}{6}$





$$\aleph$$