Question

Considerando $f(x)= \sqrt{49-x^2}$, calcule:

a) $f'(x)$

b) $4 \int\limits^7_0 {f(x)} \, dx$

Answer 1

Olá


$\displaystyle \mathsf{y= \sqrt{49-x^2} }$

A)

Usando a propriedade de derivação para raiz quadrada: $\displaystyle \mathsf{ \frac{u'}{2 \sqrt{u} } }$


$\displaystyle \mathsf{y'= \frac{-\diagup\!\!\!\!2x}{\diagup\!\!\!\!2 \sqrt{49-x^2} } }\\\\\\\boxed{\mathsf{y'= \frac{-x}{ \sqrt{49-x^2} } }}$



B)

$\displaystyle \mathsf{4 \int\limits^7_0 { \sqrt{49-x^2} } \, dx }$


Integrando pelo método da substituição trigonométrica

$\mathsf{ \sqrt{a-x^2} \qquad\Longrightarrow \qquad x=a\cdot sin\theta\qquad\Longrightarrow \qquad dx=a\cdot cos\theta}$

$\displaystyle \mathsf{4 \int\limits^7_0 { \sqrt{49-(7\cdot sen\theta)^2} } ~\cdot ~7cos\theta d\theta }\\\\\\\\\mathsf{4 \int\limits^7_0 { \sqrt{49-49sen^2\theta} } ~\cdot ~7cos\theta d\theta }\\\\\\\\\mathsf{4 \int\limits^7_0 { \sqrt{49\cdot (1- sen^2\theta}) } ~\cdot ~7cos\theta d\theta }\\\\\\\mathsf{4 \int\limits^7_0 { \sqrt{49}\cdot \sqrt{1- sen^2\theta} } ~\cdot ~7cos\theta d\theta }\\\\\\\text{Usando da identidate trigonometrica}\\\\\mathsf{sen^2x+cos^2x=1}\\\\\text{Isolando o cosseno}$


$\displaystyle \mathsf{cos^2x=1-sen^2x}\\\\\text{Substituindo na integral}\\\\\\\mathsf{4 \int\limits^7_0 { 7\cdot \sqrt{cos^2\theta} } ~\cdot ~7cos\theta d\theta }\\\\\\\mathsf{4\cdot 7\cdot 7 \int\limits^7_0 { cos\theta } ~\cdot ~cos\theta d\theta }\\\\\\\mathsf{196\int\limits^7_0 {cos^2\theta d\theta}}\\\\\\\text{Usando outra identidade trigonometrica:}\\\\\\\mathsf{cos^2x= \frac{1+cos(2x)}{2} }\\\\\\\text{Substituindo}\\\\\\\mathsf{196\int\limits^7_0 {\frac{1+cos(2\theta)}{2} d\theta}}$


$\displaystyle \mathsf{ \frac{196}{2} \int\limits^7_0 {1+cos(2\theta) d\theta}}\\\\\\\mathsf{98 \int\limits^7_0 {1+cos(2\theta) d\theta}\qquad\qquad\qquad \qquad \qquad \boxed{\int cos(\alpha x)dx= \frac{sen(\alpha x)}{\alpha} }}\\\\\\\mathsf{98\left(\theta +} \frac{sen(2\theta)}{2}\right)$


Não podemos substituir os limites inferiores e superioes da integral, pois não estamos na variável original.
Então temos que voltar para variável x.

No inicio fizemos a substituição

x = a senθ
a = 7
x = 7 senθ


$\displaystyle \mathsf{sen\theta= \frac{x}{7} }\\\\\\\boxed{\mathsf{\theta= arcsen\left( \frac{x}{7} \right)}}$


Substituindo no resultado que chegamos.

$\displaystyle \mathsf{98\left [ arcsen\left( \frac{x}{7} \right) +} \frac{1}{2}\cdot sen \left(2arcsen\left( \frac{x}{7} \right)\right)\right]$


Agora podemos substituir os limites.

$\displaystyle \mathsf{\left(98\left [ arcsen\left( \frac{x}{7} \right) +} \frac{1}{2}\cdot sen \left(2arcsen\left( \frac{x}{7} \right)\right)\right]\right)\bigg|^7_0$

$\displaystyle \mathsf{98\left [ arcsen\left( \frac{7}{7} \right) +} \frac{1}{2}\cdot sen \left(2arcsen\left( \frac{7}{7} \right)\right)\right]~-~0$

$\displaystyle \mathsf{98\left [\underbrace{ arcsen\left( 1 \right)}_{= \frac{\pi}{2} } +} \underbrace{\frac{1}{2}\cdot sen \left(2arcsen\left( 1 \right)\right)}_{=0}\right]$

$\displaystyle \mathsf{98\cdot \frac{\pi}{2} }\\\\\\\boxed{\mathsf{49\pi}}$




Dúvidas? Deixe nos comentários.




$\mathsf{AvengerCrawl\left(\smile \!\!\!\!\!\!\!^{'~'}\right)}$