Question

Calcule a integral definida abaixo:
$\int\limits^2_0 { 2x^{2} \sqrt{ x^{3}+1 } } \, dx$

Answer 1

É dada a integral: $\displaystyle I=\int_0^2 2x^2\sqrt{x^3+1}\,dx=2\int_0^2 x^2\sqrt{x^3+1}\,dx$

Vamos fazer uma substituição. Seja $u=x^3+1\Longrightarrow du=3x^2dx\Longrightarrow x^2dx=\dfrac{du}{3}$. Redefinindo os limites de integração:

Se x=0: $u=0^3+1=0+1\Longrightarrow u=1$
Se x=2: $u=2^3+1=8+1\Longrightarrow u=9$

Agora, substituindo na integral:

$\displaystyle I=2\int_0^2 x^2\sqrt{x^3+1}\,dx\\\\ I=2\int_0^2 \sqrt{x^3+1}\cdot x^2\,dx\\\\ I=2\int_1^9 \sqrt{u}\,\dfrac{du}{3}\\\\ I=\dfrac{2}{3}\int_1^9 u^{\frac 1 2}\,du\\\\ I=\dfrac{2}{3}\cdot\left[\dfrac{u^{\frac 1 2+1}}{\frac 1 2 +1}\right]^{9}_{1}\\\\ I=\dfrac{2}{3}\cdot\left[\dfrac{u^{\frac 3 2}}{\frac 3 2}\right]^{9}_{1}\\\\ I=\dfrac{4}{9}\cdot[u^{\frac 3 2}]^{9}_{1}\\\\ I=\dfrac{4}{9}\cdot[9^{\frac 3 2}-1^{\frac 3 2}]\\\\ I=\dfrac{4}{9}\cdot[27-1]\\\\ I=\dfrac{4}{9}\cdot[26]\\\\ I=\dfrac{104}{9}$

Então:

$\displaystyle \boxed{\int_0^2 2x^2\sqrt{x^3+1}\,dx=\dfrac{104}{9}}$