Question

integral definida: ∫ (x− (√ 25−x²))dx (de -5 a 5)

obs: 25-x² estão ambos dentro da raíz

Answer 1

Calcular a integral definida

     $\displaystyle\int_{-5}^5 (x-\sqrt{25-x^2})\,dx\\\\\\ =\int_{-5}^5 (x-\sqrt{5^2-x^2})\,dx$


Faça uma substituição trigonométrica:

     $\begin{array}{lcl} x=5\,\mathrm{sen\,}\theta&\quad\Rightarrow\quad& \left\{\!\begin{array}{l} dx=5\cos\theta\,d\theta\\\\ \theta=\mathrm{arcsen}\!\left(\dfrac{x}{5}\right) \end{array}\right. \end{array}$

com  − π/2 ≤ θ ≤ π/2.


Além disso,

     $\sqrt{5^2-x^2}\\\\ =\sqrt{5^2-(5\,\mathrm{sen\,}\theta)^2}\\\\ =\sqrt{5^2-5^2\,\mathrm{sen^2\,}\theta}\\\\ =\sqrt{5^2\cos^2\theta}\\\\ =5\left|\cos\theta\right|\\\\ =5\cos\theta$

pois o cosseno nunca é negativo para  θ  no intervalo  de  − π/2  a  π/2.  Assim, o módulo do cosseno é o próprio cosseno.


Novos limites de integração em  θ:

     $\begin{array}{lcl} \textrm{Quando~~}x=-\,5&\quad\Rightarrow\quad&\theta=\mathrm{arcsen}\!\left(\dfrac{-\,5}{5} \right)\\\\ &&\theta=\mathrm{arcsen}(-\,1)\\\\ &&\theta=-\,\dfrac{\pi}{2} \end{array}$


     $\begin{array}{lcl} \textrm{Quando~~}x=5&\quad\Rightarrow\quad&\theta=\mathrm{arcsen}\!\left(\dfrac{5}{5} \right)\\\\ &&\theta=\mathrm{arcsen}(1)\\\\ &&\theta=\dfrac{\pi}{2} \end{array}$


Substituindo, a integral fica

     $\displaystyle=\int_{-\pi/2}^{\pi/2} (5\,\mathrm{sen\,}\theta-5\cos\theta)\cdot 5\cos\theta\,d\theta\\\\\\ =\int_{-\pi/2}^{\pi/2} (5\,\mathrm{sen\,}\theta\cdot 5\cos\theta-5\cos\theta\cdot 5\cos\theta)\,d\theta\\\\\\ =\int_{-\pi/2}^{\pi/2} (25\,\mathrm{sen\,}\theta\cdot \cos\theta-25\cos^2\theta)\,d\theta\\\\\\ =25\int_{-\pi/2}^{\pi/2}\mathrm{sen\,}\theta\cdot \cos\theta\,d\theta-25\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta$


Para  cos² θ,  use uma das identidades trigonométricas para o arco duplo:

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


e a integral fica

     $\displaystyle=25\int_{-\pi/2}^{\pi/2}\mathrm{sen\,}\theta\cdot \cos\theta\,d\theta-25\int_{-\pi/2}^{\pi/2}\frac{1}{2}\,(1+\cos2\theta)\,d\theta\\\\\\ =25\int_{-\pi/2}^{\pi/2}\mathrm{sen\,}\theta\cdot \cos\theta\,d\theta-\frac{25}{2}\int_{-\pi/2}^{\pi/2}1\,d\theta-\frac{25}{2}\int_{-\pi/2}^{\pi/2}\cos 2\theta\,d\theta$

     $=25\left(\dfrac{\mathrm{sen^2\,}\theta}{2}\right)\!\bigg|_{-\pi/2}^{\pi/2}-\dfrac{25}{2}\,(\theta)\Big|_{-\pi/2}^{\pi/2}-\dfrac{25}{2}\left(\dfrac{\mathrm{sen\,}2\theta}{2}\right)\!\bigg|_{-\pi/2}^{\pi/2}$

     $=25\left(\dfrac{\mathrm{sen^2\,}\frac{\pi}{2}}{2}-\dfrac{\mathrm{sen^2}(-\frac{\pi}{2})}{2}\right)-\dfrac{25}{2}\left(\dfrac{\pi}{2}-\big(\!-\dfrac{\pi}{2}\big) \right )-\dfrac{25}{2}\left(\dfrac{\mathrm{sen\,}2\frac{\pi}{2}}{2}-\dfrac{\mathrm{sen\,}2(-\frac{\pi}{2})}{2}\right)$

     $=25\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\dfrac{25}{2}\left(\dfrac{\pi}{2}+\dfrac{\pi}{2}\right)-\dfrac{25}{2}\left(\dfrac{0}{2}-\dfrac{0}{2}\right)\\\\\\ =-\dfrac{25}{2}\left(\dfrac{\pi}{2}+\dfrac{\pi}{2}\right)$

      $=-\,\dfrac{25\pi}{2}$    <————    esta é a resposta.


Bons estudos! :-)