Question

O valor da integral ʃ1e2xlnx²dx é

a) 1 + e²
b) 2e²
c) e²
d) 1
e) 0

Answer 1

Olá

Alternativa correta, letra A) 1+e² 

$\displaystyle \int\limits^e_1 {2xln(x^2)} \, dx \\ \\ \\2 \int\limits^e_1 {xln(x^2)} \, dx$

Temos que resolver essa integral por partes... Dada por

$\displaystyle \boxed{\mathtt{ \int udv=uv-\int vdu}}$

$\mathtt{u=ln(x^2)} ~~~~~ ~~~~~~~ ~~~~~~~~ ~~~~ ~~~~\mathtt{dv=x}\\ \\\mathtt{du= \frac{2x}{x^2} ~=~ \frac{2}{x} }~~~~ ~~~~~~~~~ ~~~~~ ~~~~~ \mathtt{v=\int xdx = \frac{x^2}{2} }$

Substituindo na formula

Para não ficar digitando a integral toda hora, vamos chama-la de I.

$\displaystyle 2\int\limits^e_1 {xln(x^2)} \, dx ~=~I \\ \\ \\ I=2\left.\left( ln(x^2) \cdot \frac{x^2}{2}~-~\int \frac{x^2}{2} \cdot \frac{2}{x} \,\right)$

Simplifica

$\displaystyle I=2\left.\left( ln(x^2) \cdot \frac{x^{2}}{2}~-~\int \frac{x^{\diagup\!\!\!\!\!2}}{\diagup\!\!\!\!2} \cdot \frac{\diagup\!\!\!\!2}{\diagup\!\!\!\!\!x} \,\right)$

$\displaystyle I=2\left.\left( ln(x^2) \cdot \frac{x^{2}}{2}~-~\int x dx\,\right)$


$\displaystyle I=2\left.\left( ln(x^2) \cdot \frac{x^{2}}{2}~-~ \frac{x^2}{2} \,\right)$

$\displaystyle I=\diagup\!\!\!\!2\cdot ln(x^2) \cdot \frac{x^{2}}{\diagup\!\!\1\!\!2}~-~ \diagup\!\!\!\!2\cdot \frac{x^2}{\diagup\!\!\!\!2} \\ \\ \\ \boxed{I=ln(x^2)\cdot x^2~-~x^2}$

$=\left.\left(\dfrac{}{}\,\mathrm{ln(x^2)-x^2}\right)\right|_{1}^{e} \\ \\ \\ =\underbrace{(\mathsf{e^2\cdot ln(e^2)})}_{=2e^2}~-~e^2)~-~(\underbrace{(\mathsf{1^2\cdot ln(1^2)})}_{=0}~-~1) \\ \\ \\ =(2e^2-e^2)~-~(0-1) \\ \\ =\boxed{e^2+1}~~~~~ ~~~~~\longleftarrow \text{Resposta}$