Question

resolva a integral ∫3-0 (2+3/raiz9-x^2)dx.

reposta correta 6+3pi/2

Answer 1

Calcular a integral definida

     $\displaystyle\int_0^3 \left(2+\frac{3}{\sqrt{9-x^2}}\right)dx\\\\\\ =\int_0^3 \left(2+\frac{3}{\sqrt{3^2-x^2}}\right)dx$


Faça uma substituição trigonométrica:

     $\begin{array}{lcl} x=3\,\mathrm{sen\,}\theta&\quad\Rightarrow\quad&\left\{ \begin{array}{l} dx=3\cos \theta\,d\theta\\\\ \theta=\mathrm{arcsen}\!\left(\dfrac{x}{3}\right) \end{array} \right. \end{array}$

com  − π/2 ≤ θ ≤ π/2.


Além disso, temos que

     $\sqrt{3^2-x^2}\\\\ =\sqrt{3^2-(3\,\mathrm{sen\,}\theta)^2}\\\\ =\sqrt{3^2-3^2\,\mathrm{sen^2\,}\theta}\\\\ =\sqrt{3^2\cdot (1-\mathrm{sen^2\,}\theta)}\\\\ =\sqrt{3^2\cos^2\theta}\\\\ =3\left|\cos\theta\right|\\\\ =3\cos\theta$

pois neste intervalo em  θ  o valor do cosseno nunca é negativo.


Novos limites de integração em  θ:

     $\begin{array}{lcl} \textrm{Quando }x=0&\quad\Rightarrow\quad&\theta=\mathrm{arcsen}\!\left(\dfrac{0}{3}\right)\\\\ &&\theta=\mathrm{arcsen\,}0\\\\ &&\theta=0 \end{array}$


     $\begin{array}{lcl} \textrm{Quando }x=3&\quad\Rightarrow\quad&\theta=\mathrm{arcsen}\!\left(\dfrac{3}{3}\right)\\\\ &&\theta=\mathrm{arcsen\,}1\\\\ &&\theta=\dfrac{\pi}{2} \end{array}$


Então, a integral fica

     $\displaystyle\int_0^{\pi/2} \left(2+\frac{3}{3\cos \theta}\right)\cdot 3\cos\theta \,d\theta\\\\\\ =\int_0^{\pi/2} \left(6\cos\theta+\frac{3\cdot 3\cos\theta}{3\cos \theta}\right)\!d\theta\\\\\\ =\int_0^{\pi/2} (6\cos\theta+3)\,d\theta\\\\\\ =(6\,\mathrm{sen\,\theta}+3\theta)\Big|_0^{\pi/2}\\\\\\ =\left(6\,\mathrm{sen\,}\frac{\pi}{2}+3\cdot \frac{\pi}{2}\right)-\left(6\,\mathrm{sen\,}0+3\cdot 0\right)\\\\\\ =\left(6\cdot 1+\frac{3\pi}{2}\right)-0$

     $=6+\dfrac{3\pi}{2}$    <————    esta é a resposta.


Bons estudos! :-)

Answer 2

RESPOSTA CORRETA PELO AVA.