Question

Calcular a integral definida de 2 até 3 com s/(s-1)³ ds
Preciso do passo a passo.

Gabarito: 7/8

Answer 1

• O resultado dessa integral definida será 7/8.

Para resolver sua integral, devemos aplicar o método da substituição simples. Chamando assim o u de s-1 . Ficando então:

$\large\displaystyle\begin{aligned} \int _2^3 \left(\frac{s}{(s-1)^3} \right) ds \Leftrightarrow \int_2^3 \left( \frac{u+1}{(u)^3}\right) du\end{aligned}$

$\large\displaystyle\begin{aligned} \int _2^3 \left(\frac{s}{(s-1)^3} \right) ds \Leftrightarrow \int_2^3 \left( \frac{1}{(u)^2}\right) du + \int_2^3 \left( \frac{1}{(u)^3}\right) du\end{aligned}$

$\large\displaystyle\begin{aligned} \int _2^3 \left(\frac{s}{(s-1)^3} \right) ds \Leftrightarrow \int_2^3 \left((u)^{-2}\right) du+ \int_2^3 \left( (u)^{-3}\right) du\end{aligned}$

$\large\displaystyle\text{ \begin{aligned} \int _2^3 \left(\frac{s}{(s-1)^3} \right) ds \Leftrightarrow \left. \left(\frac{(u)^{-2+1}}{-2+1}\right)\right|^3_2 + \left. \left(\frac{ (u)^{-3+1}}{-3+1}\right) \right|^3_2\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \int _2^3 \left(\frac{s}{(s-1)^3} \right) ds \Leftrightarrow \left. \left(\frac{(u)^{-1}}{-1}\right)\right|^3_2 + \left. \left(\frac{ (u)^{-2}}{-2}\right) \right|^3_2\end{aligned} }$

$\large\displaystyle\begin{aligned} \int _2^3 \left(\frac{s}{(s-1)^3} \right) ds \Leftrightarrow \left. \left(-\frac{1}{u}\right)\right|^3_2 + \left. \left(-\frac{ 1}{2u^2}\right) \right|^3_2\end{aligned}$

$\large\displaystyle\begin{aligned} \int _2^3 \left(\frac{s}{(s-1)^3} \right) ds \Leftrightarrow \left. \left(-\frac{1}{s-1}\right)\right|^3_2 + \left. \left(-\frac{ 1}{2(s-1)^2}\right) \right|^3_2\end{aligned}$

• Agora, aplique o Teorema Fundamental do Cálculo.

$\large\displaystyle\begin{aligned} \left. \left(-\frac{1}{s-1}\right)\right|^3_2 + \left. \left(-\frac{ 1}{2(s-1)^2}\right) \right|^3_2\Leftrightarrow \left(-\frac{1}{2} +\frac{1}{1}\right) + \left(-\frac{ 1}{8} + \left( \frac{1}{2}\right)\right)\end{aligned}$

$\large\displaystyle\begin{aligned} \left. \left(-\frac{1}{s-1}\right)\right|^3_2 + \left. \left(-\frac{ 1}{2(s-1)^2}\right) \right|^3_2\Leftrightarrow \left(\frac{1}{2} \right) + \left(\frac{ 3}{8} \right)\end{aligned}$

$\large\displaystyle\begin{aligned} \left. \left(-\frac{1}{s-1}\right)\right|^3_2 + \left. \left(-\frac{ 1}{2(s-1)^2}\right) \right|^3_2\Leftrightarrow \frac{7}{8}\end{aligned}$

Portanto, sua integral definida é igual a:

$\large\displaystyle\begin{aligned} \left. \left(-\frac{1}{s-1}\right)\right|^3_2 + \left. \left(-\frac{ 1}{2(s-1)^2}\right) \right|^3_2\Leftrightarrow \frac{7}{8}\end{aligned}$

$\large\displaystyle\text{ \begin{aligned} \therefore \boxed{\boxed{\green{\int _2^3 \left(\frac{s}{(s-1)^3} \right) ds \Leftrightarrow \frac{7}{8} }}}\end{aligned} }$

Veja mais sobre:

Integrais definidas.

$\blue{\square}$ brainly.com.br/tarefa/5048105