Question

o valor da integral
arquivo em anexo

Answer 1

É dada a seguinte integral definida:

$\displaystyle I=\int_{-1}^0x^2\sqrt{x^3+1}\,dx$

Vamos fazer uma substituição:

$u=x^3+1\Longrightarrow du=3x^2\,dx\Longrightarrow dx=\dfrac{1}{3x^2}du$

Substituindo na integral dada:

$\displaystyle I=\int_{-1}^0x^2\sqrt{x^3+1}\,dx\\\\ I=\int_{u_1}^{u_2}x^2\sqrt{u}\cdot\dfrac{1}{3x^2}du\\\\ I=\dfrac{1}{3}\int_{u_1}^{u_2}\sqrt{u}\,du\\\\ I=\dfrac{1}{3}\int_{u_1}^{u_2}u^{\frac{1}{2}}\,du\\\\ I=\dfrac{1}{3}\left[\dfrac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{u_1}^{u_2}\\\\ I=\dfrac{1}{3}\left[\dfrac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_{u_1}^{u_2}\\\\ I=\dfrac{2}{9}\left[u^{\frac{3}{2}}\right]_{u_1}^{u_2}$

Voltando à variável x:

$I=\dfrac{2}{9}\left[u^{\frac{3}{2}}\right]_{u_1}^{u_2}\\\\ I=\dfrac{2}{9}\left[(x^3+1)^{\frac{3}{2}}\right]_{-1}^{0}\\\\ I=\dfrac{2}{9}\left[(0^3+1)^{\frac{3}{2}}-((-1)^3+1)^{\frac{3}{2}}\right]\\\\ I=\dfrac{2}{9}\left[(0+1)^{\frac{3}{2}}-(-1+1)^{\frac{3}{2}}\right]\\\\ I=\dfrac{2}{9}\left[1^{\frac{3}{2}}-0^{\frac{3}{2}}\right]\\\\ I=\dfrac{2}{9}[1-0]\\\\ I=\dfrac{2}{9}\cdot1\\\\ I=\dfrac{2}{9}\\\\ \displaystyle \boxed{\int_{-1}^0x^2\sqrt{x^3+1}\,dx=\dfrac{2}{9}}\Longrightarrow \text{Letra }\bold{D}$