Ajuda para resolver Integral definida

por **rodolphogagno** » Qua Dez 01, 2010 15:16

Pessoal, alguém pode me ajudar a resolver essas questões?
$a) \int_{0}^{\frac{\Pi}{2}} sen\, 2x\, dx$
...........................................................................
$b) \int_{-2}^{0} 3w\, \sqrt[]{4-{w}^{2}}\,dw$

Alguém se manifesta? rs

por **DanielFerreira** » Qua Dez 01, 2010 17:28

Seja,
2x = u
du = 2 dx ====> dx = du/2

$\int_{0}^{\frac{\pi}{2}} sen\,u\, \frac{du}{2} =$

$\frac{1}{2}\int_{0}^{\frac{\pi}{2}} sen\,u\, du =$

$\frac{1}{2} . - cos\,u\,\int_{0}^{\frac{\pi}{2}} du =$

$\frac{- cos\,2x\,}{2}\int_{0}^{\frac{\pi}{2}} =$

$F(\frac{\pi}{2}) - F(0) =$

$\frac{ - cos\,2.\frac{\pi}{2}}{2} - \frac{- cos\,2.0}{2} =$

$\frac{ - cos\,\pi}{2} + \frac{cos\,0}{2} =$

$- \frac{- 1}{2} + \frac{1}{2} =$

por **rodolphogagno** » Qua Dez 01, 2010 17:44

O que acha da questão B meu caro?

por **DanielFerreira** » Qua Dez 01, 2010 17:49

Sai por Função Trigonométrica!

por **Moura** » Seg Dez 13, 2010 21:51

b) $u=4-{w}^{2}===>\frac{du}{dw}=-2w ===>-2wdw=du ===> dw=\frac{du}{-2w}$

$\int_{-2}^{0}3w\sqrt[]{4-{w}^{2}}dw =$ $\int_{-2}^{0}3w{u}^{\frac{1}{2}}dw =$ $\int_{-2}^{0}3w{u}^{\frac{1}{2}}\frac{du}{-2w}=$ $\int_{-2}^{0}-\frac{3}{2}{u}^{\frac{1}{2}}du=$

$-\frac{3}{2}\frac{{u}^{\frac{3}{2}}}{\frac{3}{2}}]_{-2}^3=$ $\frac{2}{3}\left(\frac{-3}{2} \right){u}^{\frac{3}{2}} du]_{-2}^3=$ $-{u}^{\frac{3}{2}}]_{-2}^3$

$(-(4-(0{)}^{2}{{)}^{\frac{3}{2}})-(-(4-(-2{)}^{2}{)}^{\frac{3}{2}}= -8$

:y: