Integral definida: ∫ (x− (√ 25−x²))dx (de -5 a 5) obs: 25-x² estão ambos dentro da raíz.... Pergunta de ideia deQueenOfPain

December 2019 1 66 Report

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

obs: 25-x² estão ambos dentro da raíz

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Lukyo
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! :-)

QueenOfPain muito obrigada!!!

Lukyo De nada. :)

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