Question

resolver a integral pelo metodo da substituição. Me ajudem​

Answer 1

Resposta:

$\int5x\sqrt{4-3x^2}\;dx=-\frac{5}{9}\,(4-3x^2)^{3/2}+C$

Explicação passo-a-passo:

Considerando $4-3x^2=u$, temos que:

$\frac{du}{dx}=\frac{d}{dx}(4-3x^2)$

$\frac{du}{dx}=\frac{d}{dx}(4)+\frac{d}{dx}(-3x^2)$

$\frac{du}{dx}=0-6x$

$\frac{du}{dx}=-6x\therefore dx=-\frac{1}{6x}\;du$

Substituindo $dx$ no integrando:

$\int5x\sqrt{4-3x^2}\;dx=\int5x\sqrt{u}*\left(-\frac{1}{6x}\right)\;du$

$\int5x\sqrt{4-3x^2}\;dx=\int-\frac{5}{6}\sqrt{u}\;du$

$\int5x\sqrt{4-3x^2}\;dx=-\frac{5}{6}\int\sqrt{u}\;du$

$\int5x\sqrt{4-3x^2}\;dx=-\frac{5}{6}\int u^{1/2}\;du$

$\int5x\sqrt{4-3x^2}\;dx=-\frac{5}{6}*\frac{u^{1/2+1}}{1/2+1}+C$

$\int5x\sqrt{4-3x^2}\;dx=-\frac{5}{6}*\frac{u^{3/2}}{3/2}+C$

$\int5x\sqrt{4-3x^2}\;dx=-\frac{5}{6}*\frac{2}{3}*u^{3/2}+C$

$\int5x\sqrt{4-3x^2}\;dx=-\frac{5}{9}\,u^{3/2}+C$

$\int5x\sqrt{4-3x^2}\;dx=-\frac{5}{9}\,(4-3x^2)^{3/2}+C$