Question

O valor da integral ∫(intervalo de e^3 até e^2 ) x ln xdx

Answer 1

Calcular a integral definida:

     $\displaystyle\int_{e^2}^{e^3}x\ln x\,dx$


Vamos fazer a integração por partes. Computando a integral indefinida primeiro:

     $\displaystyle\int x\ln x\,dx\\\\\\ =\int (\ln x)\cdot x\,dx\\\\\\\ \begin{array}{lcl} u=\ln x&\quad\Rightarrow\quad&du=\dfrac{1}{x}\,dx\\\\ dv=x\,dx&\quad\Leftarrow\quad&v=\dfrac{x^2}{2} \end{array}$


     $\displaystyle\int u\,dv=uv-\int v\,du\\\\\\ \int (\ln x)\cdot x\,dx=(\ln x)\cdot \frac{x^2}{2}-\int \frac{x^2}{2}\cdot \frac{1}{x}\,dx\\\\\\ \int x\ln x\,dx=\frac{x^2\ln\,x}{2}-\frac{1}{2}\int \frac{x^2}{x}\,dx\\\\\\ \int x\ln x\,dx=\frac{x^2\ln\,x}{2}-\frac{1}{2}\int x\,dx$

     $\displaystyle\int x\ln x\,dx=\frac{x^2\ln\,x}{2}-\frac{1}{2}\cdot \frac{x^2}{2}+C\\\\\\ \int x\ln x\,dx=\frac{x^2\ln\,x}{2}-\frac{x^2}{4}+C$


De posse da primitiva da função, podemos computar a integral definida, usando o Teorema Fundamental do Cálculo:

     $\displaystyle\int_{e^2}^{e^3} x\ln x\,dx\\\\\\ =\bigg(\frac{x^2\ln\,x}{2}-\frac{x^2}{4}\bigg)\bigg|_{e^2}^{e^3}\\\\\\ =\bigg(\frac{(e^3)^2\ln(e^3)}{2}-\frac{(e^3)^2}{4}\bigg)-\bigg(\frac{(e^2)^2\ln(e^2)}{2}-\frac{(e^2)^2}{4}\bigg)\\\\\\ =\bigg(\frac{e^6\cdot 3}{2}-\frac{e^6}{4}\bigg)-\bigg(\frac{e^4\cdot 2}{2}-\frac{e^4}{4}\bigg)\\\\\\ =e^6\cdot \bigg(\frac{3}{2}-\frac{1}{4}\bigg)-e^4\cdot \bigg(\frac{2}{2}-\frac{1}{4}\bigg)$

     $=e^6\cdot \bigg(\dfrac{6}{4}-\dfrac{1}{4}\bigg)-e^4\cdot \bigg(\dfrac{4}{4}-\dfrac{1}{4}\bigg)\\\\\\ =e^6\cdot \dfrac{5}{4}-e^4\cdot \dfrac{3}{4}$

     $=\dfrac{5e^6}{4}-\dfrac{3e^4}{4}\approx 463,\!34$    <————    esta é a resposta.


Bons estudos! :-)

Answer 2

Resposta:

Resposta correta 463,34

Explicação passo a passo:

corrigido pelo ava