Question

Marque a alternativa que corresponde ao valor da integral a seguir: integral 3n/2 n/2 (sen(3x)-cos(3x))dx :
a) 1
b) -1
c) 0
d) -2/3 (Essa é a CORRETA)
e) 2/3

Answer 1

Sabemos que:

$\int\limits^b_a {Sen(nx)} \, dx = - \frac{Cos(nx)}{n} |(a,b)$

E que:

$\int\limits^b_a {Cos(nx)} \, dx = \frac{Sen(nx)}{n} |(a,b)$

Então teremos:

$\int\limits^b_a {Sen(3x)-Cos(3x)} \, dx = - \frac{Cos(3x)}{3} - \frac{Sen(3x)}{3} |( \frac{ \pi }{2} , \frac{3 \pi }{2} ) \\ \\ = -Cos(9 \pi /2)/3 -Sen(9 \pi /2)/3 -[-Cos(3 \pi /2)/3 -Sen(3 \pi /2)/3] \\ \\ = -0 -(1)/3 - [ -0-(-1/3)] \\ \\ = -1/3 - (1/3) \\ \\ = -2/3$
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Observação:

Para calcular os valores de 9π/2 e 3π/2

Substitue π por 180.

9π/2 = 9×180/2 = 810°

3π/2 = 3×180/2 = 270°
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Answer 2

$\int\limits^{\frac{3 \pi }{2} }_{ \frac{ \pi }{2}} {(sen(3x)-cos(3x))} \, dx$

$\int\limits^{\frac{3 \pi }{2} }_{ \frac{ \pi }{2}} {sen(3x)} \, dx -\int\limits^{\frac{3 \pi }{2} }_{ \frac{ \pi }{2}} {cos(3x)} \, dx$

u = 3x                       w = 3x
du = 3dx                   dw = 3dx

$\int\limits { \frac{1}{3}sen(u) } \, du- \int\limits { \frac{1}{3}cos(w) } \, dw$

$- \dfrac{1}{3}cos(u)- \dfrac{1}{3}sen(w)$

$-\dfrac{1}{3}cos(3x)- \dfrac{1}{3}sen(3x) \Bigg|_{ \dfrac{ \pi }{2}} ^{ \dfrac{3 \pi }{2} }$

$-\dfrac{1}{3}[sen(3x)+cos(3x)] \Bigg|_{ \dfrac{ \pi }{2}} ^{ \dfrac{3 \pi }{2} }$

$-\dfrac{1}{3}[sen(3. \dfrac{3 \pi }{2} )+cos(3. \dfrac{3 \pi }2} )]+\dfrac{1}{3}[sen(\dfrac{3 \pi }{2} )+cos(3. \dfrac{\pi }2} )]$

$-\dfrac{1}{3} (1+0)+ \dfrac{1}{3}(-1+0)\\\\= -\dfrac{1}{3}- \dfrac{1}{3}\\\\= -\dfrac{2}{3}$

LETRA D