Question

O valor da integral 6X²(X³+2)² no intervalo de 0 e 1:
a:0
b:12
c:6
d:38/3
e:1/13

Answer 1

Olá

Resposta correta, letra d)

$\displaystyle \int_{0}^{1} 6x^2(x^3+2)^2dx \\ \\ \\\text{tira a constante da integral}\\\\\\6\int_{0}^{1} x^2(x^3+2)^2dx\\\\\\ \text{Temos que resolver por substituicao 'udu'} \\ \\ \\ \mathsf{u=x^3+2} \\ \\ \mathsf{du=3x^2dx} \\ \\ \text{passa o 3 para o outro lado dividindo} \\ \\ \mathsf{\frac{1}{3} du=x^2dx} \\ \\ \\ \text{substituindo na integral} \\ \\ \\ 6\int_{0}^{1} u^2\cdot \frac{1}{3} du \\ \\ \\ \text{tira a constante } \frac{1}{3} \text{ de 'dentro' da integral}$

$\displaystyle \frac{1}{\diagup\!\!\!\!3} \cdot \diagup\!\!\!\!6\int_{0}^{1} u^2du \\ \\ \\ \\ 2\int_{0}^{1} u^2 du \\ \\ \\ =2\cdot \left.\left(\dfrac{u^{2+1}}{2+1}\,\right)$

$\displaystyle =2\cdot \frac{u^3}{3} \\ \\ \\ \text{Voltando com }x^2+2\text{ no lugar do 'u'} \\ \\ \\=2\cdot \left.\left(\dfrac{(x^3+2)^3}{3}\,\right)\right|_{0}^{1} \\ \\ \\ =2\cdot \left.\left(\dfrac{(1^3+2)^3}{3}~~~~ ~-~~~~~ \frac{(0^3+2)^3}{3} \,\right)$

$\displaystyle =2\cdot \left.\left(\dfrac{(3)^3}{3}~~ ~-~~~ \frac{(2)^3}{3} \,\right)$

$\displaystyle =2\cdot \left.\left(\dfrac{27}{3}~~ ~-~~~ \frac{8}{3} \,\right)$

$\displaystyle =2\cdot \left.\left(\dfrac{19}{3} \,\right)$

$\displaystyle =~\boxed{\dfrac{38}{3}}$