Question

R e a região sob a curva y =√9-5x² no intervalo 0≤x≤1

Answer 1

$\boxed{\boxed{ Area=\int\limits^1_0 { \sqrt{9-5x^2} } \, dx }}$

quando vc tem algo do tipo
$\boxed{ \sqrt{a^2-bx^2}}$

faça a substituição
$\boxed{x= \frac{ \sqrt{a} }{ \sqrt{b} }*sen(t) }$

aplicando isso
$x= \frac{ \sqrt{9} }{ \sqrt{5} } *sen(t)= \frac{3 }{ \sqrt{5} } *sen(t)\\\\dx= \frac{3}{\sqrt{5}}*cos(t)*dt$

temos
$\int\limits {[\sqrt{9-5*(\frac{3 }{ \sqrt{5} } *sen(t))^2}}]* \frac{3}{\sqrt{5}}*cos(t)*dt \\\\= \frac{3}{\sqrt{5}} \int\limits{\sqrt{9-5*( \frac{3^2}{\sqrt{5}^2} )*sen^2(t)}}*cos(t)*dt\\\\= \frac{3}{\sqrt{5}} \int\limits {\sqrt{9-5* \frac{9}{5} *sen^2(t)}}*cos(t)*dt\\\\\frac{3}{\sqrt{5}} \int\limits {\sqrt{9-9 *sen^2(t)}}*cos(t)*dt\\\\\frac{3}{\sqrt{5}} \int\limits {\sqrt{9*(1- sen^2(t))}}*cos(t)*dt$

$\frac{3}{\sqrt{5}} \int\limits {\sqrt{9*(1- sen^2(t))}}*cos(t)*dt\\\\\frac{3}{\sqrt{5}} \int\limits {\sqrt{9}*\sqrt{(1- sen^2(t))}}*cos(t)*dt\\\\\frac{3}{\sqrt{5}} \sqrt{9}\int\limits {\sqrt{1- sen^2(t)}}*cos(t)*dt\\\\\boxed{ \frac{9}{\sqrt{5}} \int\limits\sqrt{1-sen^2(t)}*cos(t)*dt }$

usando um pouco de trigonometria

$sen^2(x)+cos^2(x)=1\\\\cos^2(x)=1-sen^2(x)$

então temos

$\frac{9}{\sqrt{5}} \int\limits \sqrt{cos^2(t)}*cos(t)*dt }\\\\ \frac{9}{\sqrt{5}} \int\limits cos(t)*cos(t)*dt }\\\\\boxed{ \frac{9}{\sqrt{5}} \int\limits cos^2(t) }$

como 

$cos^2(t)= \frac{1+cos(2t)}{2}$

a integral fica
$\frac{9}{\sqrt{5}} \int\limits \frac{1+cos(2t)}{2} }\\\\\\ \frac{9}{\sqrt{5}}( \int \frac{1}{2}dt +\int \frac{cos(2t)}{2} dt)\\\\\boxed{\frac{9}{2\sqrt{5}}*\left [ t+ \int \frac{cos(2t)}{2} \right ]}$

fazendo a substituição 
$u = 2t\\\\du=2dt\to\frac{du}{2}=dt$

$\int cos(u) * \frac{du}{2} = \frac{sen(u)}{2}= \frac{sen(2t)}{2}= \frac{2*sen(t)*cos(t)}{2} =sen(t)*cos(t)$

temos a expressão
$\frac{9}{2\sqrt{5}}*\left [ t+ sen(t)*cos(t)\right ]$

voltando para a variavel x
$x= \frac{ 3 }{ \sqrt{5 }}*sen(t) }\\\\ \boxed{\boxed{\frac{x\sqrt{5}}{3}=sen(t) }} \\\\\\\boxed{\boxed{arcsen(\frac{\sqrt{5}x}{3} )=t}}$

.$cos(t)= \sqrt{1-sen^2(t)} \\\\cos(t)=\sqrt{1-(\frac{x\sqrt{5}}{3})^2)}\\\\\boxed{\boxed{cos(t)=\sqrt{1- \frac{5x^2}{9} }}}$


então o resultado da integral ficou
vou escrever como primitiva

$\int\limits { \sqrt{9-5x^2} } \, dx \\\\ \boxed{\boxed{F(x)=\frac{9}{2\sqrt{5}}*\left [ arcsen\left (\frac{x\sqrt{5}}{3} \right )+ \left (\frac{x\sqrt{5}}{3} \right )*\sqrt{1-\frac{5x^2}{9}}\right ]}}$

o resultado da integral definida de 0 a 1 será
Area =F(1) - F(0)
Area =2,692618 - 0
Area = 2,692618 ua