Question

O valor da integral é:

Answer 1

Vamos resolver primeiro a integral de dentro, a qual está entre colchetes:

$I_1=\displaystyle\int^{x}_{x^2}dy=[y]^{x}_{x^2}\\\\ I_1=x-x^2$

Agora podemos resolver a integral "de fora":

$I_2=\displaystyle\int^{1}_{0}\left[\displaystyle\int_{x^2}^{x}dy\right]dx =\displaystyle\int^{1}_{0}(x-x^2)dx\\\\ I_2=\displaystyle\int^{1}_{0}x\,dx-\displaystyle\int^{1}_{0}x^2\,dx\\\\ I_2=\left[\dfrac{x^2}{2}\right]^{1}_{0}-\left[\dfrac{x^3}{3}\right]^{1}_{0}\\\\ I_2=\left[\left(\dfrac{1^2}{2}\right)-\left(\dfrac{0^2}{2}\right)\right]-\left[\left(\dfrac{1^3}{3}\right)-\left(\dfrac{0^3}{3}\right)\right]$

$I_2=\left[\left(\dfrac{1}{2}\right)-\left(\dfrac{0}{2}\right)\right]-\left[\left(\dfrac{1}{3}\right)-\left(\dfrac{0}{3}\right)\right]\\\\ I_2=\left[\left(\dfrac{1}{2}\right)-0\right]-\left[\left(\dfrac{1}{3}\right)-0\right]\\\\ I_2=\left(\dfrac{1}{2}\right)-\left(\dfrac{1}{3}\right)=\dfrac{3-2}{6}\\\\ \boxed{I_2=\dfrac{1}{6}}\Longrightarrow\text{Letra }\bold{D}$

Answer 2

Resposta:

reposta: 1/6 corrigido pelo AVA.

Explicação passo a passo: