Question

6) Calcule a seguinte integral definida:

$\int\limits^1_0 { \sqrt{1-x^2} } \, dx$

Answer 1

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Calcular a integral definida:

     $\mathsf{\displaystyle\int_0^1\!\sqrt{1-x^2}\,dx}$


Usamos substituição trigonométrica:

     $\mathsf{x=sen\,\theta}\quad \Rightarrow \quad\left\{ \begin{array}{l} \mathsf{dx=cos\,\theta\,d\theta}\\\\ \mathsf{\theta=arcsen(x)} \end{array} \right.$

com  − π/2 ≤ θ ≤ π/2.


Além disso, temos que

     $\mathsf{\sqrt{1-x^2}}\\\\ =\mathsf{\sqrt{1-sen^2\,\theta}}\\\\ =\mathsf{\sqrt{cos^2\,\theta}}\\\\ =\mathsf{|cos\,\theta|}\\\\ =\mathsf{cos\,\theta}$

Nesse caso o módulo é o próprio cosseno, pois  cos θ  nunca é negativo para  − π/2 ≤ θ ≤ π/2.


Novos limites de integração em  θ:

     Quando  x = 0,   θ = arcsen(0) = 0

     Quando  x = 1,   θ = arcsen(1) = π/2


Então, a integral fica

     $=\mathsf{\displaystyle\int_0^{\pi/2}cos\,\theta\cdot cos\,\theta\,d\theta}\\\\\\ =\mathsf{\displaystyle\int_0^{\pi/2}cos^2\,\theta\,d\theta}$


Use uma das identidades trigonométricas para o cosseno do arco duplo:

     •   cos² θ = (1/2) · (1 + cos 2θ)


e a integral definida fica

     $=\mathsf{\displaystyle\int_0^{\pi/2}\frac{1}{2}\cdot (1+cos\,2\theta)\,d\theta}\\\\\\ =\mathsf{\displaystyle\frac{1}{2}\int_0^{\pi/2}1\,d\theta+\frac{1}{2}\int_0^{\pi/2}\!cos\,2\theta\,d\theta}\\\\\\ =\mathsf{\displaystyle\frac{1}{2}\int_0^{\pi/2}1\,d\theta+\frac{1}{2}\int_0^{\pi/2}\!cos\,2\theta\cdot \frac{1}{2}\cdot 2\,d\theta}\\\\\\ =\mathsf{\displaystyle\frac{1}{2}\int_0^{\pi/2}1\,d\theta+\frac{1}{4}\int_0^{\pi/2}\!cos\,2\theta\cdot 2\,d\theta\qquad\quad(i)}$


Para a 2ª integral acima, faça outra substituição:

     2θ = u     ⇒     2 dθ = du


Novos limites de integração em  u:

     Quando  θ = 0,    u = 2 · 0 = 0

     Quando  θ = π/2,    u = 2 · π/2 = π


e a integral  (i)  fica

     $=\mathsf{\displaystyle\frac{1}{2}\int_0^{\pi/2}1\,d\theta+\frac{1}{4}\int_0^\pi\!cos\,u\,du}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \theta\big|_0^{\pi/2}+\dfrac{1}{4}\cdot sen\,u\big|_0^\pi}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \left(\dfrac{\pi}{2}-0\right)+\dfrac{1}{4}\cdot (sen\,\pi-sen\,0)}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}+0}$

     $=\mathsf{\dfrac{\pi}{4}}$   <————   esta é a resposta.


Bons estudos! :-)

Answer 2

$\mathsf{x=1sen(\theta)\to\,dx=cos(\theta)d \theta}\\\mathsf{\sqrt{1-{x}^{2}}=cos(\theta)}\\\mathsf{x=0\to\,\theta=0}\\\mathsf{x=1\to\,\theta=\dfrac{\pi}{2}}$

$\displaystyle\mathsf{\int\limits_{0}^{1}\sqrt{1-{x}^{2}}dx = \int \limits_{0}^{ \frac{\pi}{2}} {cos}^{2} \theta \: d \theta }$

$\mathsf{ \dfrac{1}{2}(\theta+sen(2\theta) )\Bigg | _{0}^{ \dfrac{\pi}{2}}=\dfrac{1}{2}. \dfrac{\pi}{2} =\dfrac{\pi}{4}}$