Question

Calcule a seguinte integral definida de ∫sen(2x)dx no intervalo de pi/2 a 0

Answer 1

Temos a seguinte integral:

$\int\limits^0_\frac{\pi}{2}{\sin(2x)} \, dx$

Vamos fazer a seguinte substituição:

u = 2x

du/dx = 2 ⇒ dx = du/2

Note que para x variando entre π/2 e 0, temos que u varia de π a 0. Assim a integral fica

$\int\limits^0_\frac{\pi}{2}{\sin(2x)} \, dx=\\\\\int\limits^0_\pi{\sin(u)} \, \frac{du}{2}=\\\\ \frac{1}{2}\int\limits^0_\pi{\sin(u)} \, du=\\\\ \frac{1}{2}*[(-\cos0)-(-\cos\pi)]=\\\\ \frac{1}{2}*[-\cos0+\cos\pi]=\\\\ \frac{1}{2}*[-(1)+(-1)]=\\\\ \frac{1}{2}*[-1-1]=\\\\ \frac{1}{2}*[-2]=\\\\-1$

Answer 2

Resolução da questão, veja:

Vamos primeiramente reescrever esta integral de maneira a simplificar nossos cálculos, veja:

$\mathsf{\displaystyle\int_{\frac{\pi}{2}}^{0}~sen(2x)~dx}}}=\mathsf{\displaystyle\int_{0}^{\frac{\pi}{2}}~\dfrac{-cos(2x)}{2}~dx}}}\\\\\\\\\ \mathsf{-cos~\dfrac{(2~\cdot~0)}{2}}-\mathsf{\left(-cos~\frac{\frac{2\pi}{2}}{2}}\right)}}\\\\\\\\ \mathsf{-cos~\dfrac{(0)}{2}}-\mathsf{\left(-cos~\dfrac{\pi}{2}}\right)}}\\\\\\\\\ \mathsf{-1-0}}}}}\\\\\\\\\ \Large{\boxed{\boxed{\boxed{\boxed{\mathbf{\displaystyle\int_{\frac{\pi}{2}}^{0}~sen(2x)~dx}=-1.}}}}}}}}}}}}}}}}~~\checkmark}}$

Espero que te ajude. '-'