Question

Analise a veracidade da seguinte afirmação:

A área compreendida entre f(x) = 1/x e g(x)= -x/5+6/5 é igual a 6/5-ln(5).

Preciso entender como desenvolver

Answer 1

Boa noite!

Solução!

Já que trata-se de um calculo de área compreendido entre duas funcões,precisamos encontrar os intervalos de integração igualando as funções.


$f(x)= \dfrac{1}{x} \\\\\\\\ g(x)= -\dfrac{x}{5}+ \dfrac{6}{5}\\\\\\\\ \dfrac{1}{x} =-\dfrac{x}{5}+ \dfrac{6}{5}\\\\\\\ Reescrevendo~~a~~func\~ao~~do~~lado~~direito!\\\\\\\ \dfrac{1}{x} =-\dfrac{x+6}{5}\\\\\\\ 5=- x^{2} +6x$

$x^{2} -6x+5=0\\\\\\\ x= \dfrac{-(-6)\pm \sqrt{(-6)^{2}-4.1+5 } }{2.1} \\\\\\\\\ x= \dfrac{6\pm \sqrt{36-20} }{2} \\\\\\\\\ x= \dfrac{6\pm \sqrt{16} }{2} \\\\\\\\\ x= \dfrac{6\pm 4 }{2} \\\\\\\\\ x1= \dfrac{6+4}{2}= \frac{10}{2}=5\\\\\\\ x1= \dfrac{6-4}{2}= \frac{2}{2}=1\\\\\\\ Intervalo= [1,5]$


$A=\displaystyle \int_{1}^{5} \frac{1}{x}dx+ A=\displaystyle\int_{1}^{5} -\frac{x}{5}+ \frac{6}{5}dx \\\\\\\\ A= ln(x)-\dfrac{ x^{2} }{10}+ \dfrac{6x}{5}-ln(x)-\dfrac{ x^{2} }{10}+ \dfrac{6x}{5} \bigg|_{1}^{5}$


$A= ln(5)-\dfrac{ 5^{2} }{10}+ \dfrac{6.5}{5}-ln(1)-\dfrac{ 1^{2} }{10}+ \dfrac{6.1}{5} \\\\\\\\ A= (ln(5)-\dfrac{ 25}{10}+ 6)-(ln(1)-\dfrac{ 1}{10}+ \dfrac{6}{5}) \\\\\\\\ A= (\dfrac{10ln(5)-25+60 }{10})-(\dfrac{10ln(1)-1+12}{10}}) \\\\\\\\ A= (\dfrac{10ln(5)+35 }{10})-(\dfrac{10ln(1)+11}{10}}) \\\\\\\\ A= (\dfrac{10ln(5)+35 }{10}-\dfrac{10ln(1)-11}{10}}) \\\\\\\\ A= (\dfrac{16+35 }{10}-\dfrac{0-11}{10}})$

$A= (\dfrac{51-11}{10}) \\\\\\\\ A= \dfrac{40}{10})\\\\\\ \boxed{A= 4~u.a}$


$\boxed{Resposta: A~~afirmac\~ao~~e~~falsa}$


Boa noite!

Bons estudos!