• Matéria: Matemática
  • Autor: pedroleonel83
  • Perguntado 5 anos atrás

me ajudem a resolver esta integral?​

Anexos:

Respostas

respondido por: CyberKirito
1

\displaystyle\sf\int_{-1}^{0} x^2\sqrt{1+x^3}~dx=\dfrac{1}{3}\int_{-1}^{0}3x^2\sqrt{1+x^3}~dx\\\sf fac_{\!\!,}a~u=1+x^3\\\sf du=3x^2~dx\\\sf se~x=-1\implies u=0\\\sf se~x=0\implies u=1

\displaystyle\sf\int_{-1}^{0}\sqrt{1+x^3}~dx=\dfrac{1}{3}\int_{0}^{1}\sqrt{u}~du=\dfrac{1}{3}\cdot\dfrac{2}{3}u^{\frac{3}{2}}\Bigg|_{0}^{1}\\\sf\dfrac{2}{9}u^{\frac{3}{2}}\Bigg|_{0}^{1}=\dfrac{2}{9}\cdot1^{\frac{3}{2}}=\dfrac{2}{9}\green\checkmark


pedroleonel83: Obrigado Rubens
CyberKirito: De nada ^^
analuisa131512: pedro cadê vc
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