Question

qual a resposta da integral definida ?

Answer 1

$\int\limits^{e^2}_e { \frac{ln(x)}{x} } \, dx$

fazendo a substituição
$\boxed{u = ln(x)}\\\\du = \frac{1}{x}*dx \\\\\boxed{x*du=dx}$

ficando com 
$\int\limits^{e^2}_e { \frac{u}{x}*x } \, du\\\\= \int\limits^{e^2}_e { u } \, du= \left | \frac{u^2}{2} \right |^{e^2}_e= \frac{[ln(e^2)]^2-[ln(e)]^2}{2}$

como 
ln(e) =1
ln(e²) = ln(e) + ln(e) = 2ln(e) = 2

temos
$\boxed{\boxed{ \int\limits^{e^2}_e { \frac{ln(x)}{x} } \, dx= \frac{2^2-1^2}{2}= \frac{3}{2} }}$

Answer 2

✅ Após resolver os cálculos, concluímos que a integral definida da referida função no respectivo intervalo de integração é:

  $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \int_{e}^{e^{2}} \frac{\ln(x)}{x}\,dx = \frac{3}{2}\:u\cdot a\:\:\:}}\end{gathered} }$    

 

Seja a integral definida:

                   $\Large\displaystyle\text{ \begin{gathered}\tt \int_{e}^{e^{2}} \frac{\ln(x)}{x}\,dx\end{gathered} }$

Para resolver esta integral devemos utilizar o método da substituição. Para isso devemos:

• Nomear o numerador da função:

                   $\Large\displaystyle\begin{gathered}\tt u = \ln(x)\end{gathered}$

• Derivar "u" em termos de "x":

                     $\Large\displaystyle\begin{gathered}\tt \frac{du}{dx} = \frac{1}{x}\end{gathered}$

                     $\Large\displaystyle\begin{gathered}\tt du = \frac{1}{x}\,dx\end{gathered}$

• Realizar a substituição, desenvolver e simplificar os cálculos;

       $\Large\displaystyle\text{ \begin{gathered}\tt \int_{e}^{e^{2}} u\cdot du = \Bigg(\frac{u^{1 + 1}}{1 + 1}\Bigg)\Bigg|_ {e}^{e^{2}}\end{gathered} }$

                              $\Large\displaystyle\text{ \begin{gathered}\tt = \Bigg(\frac{u^{2}}{2}\Bigg)\Bigg|_{e}^{e^{2}}\end{gathered} }$

                              $\Large\displaystyle\text{ \begin{gathered}\tt = \Bigg\{\frac{\left[\ln(e^{2})\right]^{2}}{2} + c\Bigg\} - \Bigg\{\frac{\left[\ln(e)\right]^{2}}{2} + c\Bigg\}\end{gathered} }$

                              $\Large\displaystyle\text{ \begin{gathered}\tt = \Bigg\{\frac{\left[\ln(e) + \ln(e)\right]^{2}}{2} + c\Bigg\} - \Bigg\{\frac{1^{2}}{2} + c\Bigg\}\end{gathered} }$

                              $\Large\displaystyle\text{ \begin{gathered}\tt = \Bigg\{\frac{\left[2\cdot \ln(e)\right]^{2}}{2} + c\Bigg\} - \Bigg\{\frac{1}{2} + c\Bigg\}\end{gathered} }$

                              $\Large\displaystyle\text{ \begin{gathered}\tt = \Bigg\{\frac{\left[2\cdot1\right]^{2}}{2} + c\Bigg\} - \Bigg\{\frac{1}{2} + c\Bigg\}\end{gathered} }$

                              $\Large\displaystyle\text{ \begin{gathered}\tt = \Bigg\{\frac{2^{2}}{2} + c\Bigg\} - \Bigg\{\frac{1}{2} + c\Bigg\}\end{gathered} }$

                              $\Large\displaystyle\begin{gathered}\tt = \frac{4}{2} + c - \frac{1}{2} - c\end{gathered}$

                              $\Large\displaystyle\begin{gathered}\tt = \frac{4}{2} - \frac{1}{2}\end{gathered}$

                              $\Large\displaystyle\begin{gathered}\tt = \frac{4 - 1}{2}\end{gathered}$

                              $\Large\displaystyle\begin{gathered}\tt = \frac{3}{2}\end{gathered}$

✅ Portanto, a integral definida procurada é:

       $\Large\displaystyle\text{ \begin{gathered}\tt \int_{e}^{e^{2}} \frac{\ln(x)}{x}\,dx = \frac{3}{2}\:u\cdot a\end{gathered} }$

     

Saiba mais:

1. https://brainly.com.br/tarefa/8202825
2. https://brainly.com.br/tarefa/4341918
3. https://brainly.com.br/tarefa/36147694
4. https://brainly.com.br/tarefa/2409823

Solução gráfica: