Question

O resultado da integra lé:

Answer 1

Oi . 
Segue o resultado :)

$\int\limits^2_{-2} {3 \sqrt{3x+7}} \, dx \ \ \ \ \ u=3x+7 \ \ \ \frac{du}{dx}=3 \ \ \ \ dx= \frac{du}{3} \\ \\ \int\limits^2_{-2} {3 \sqrt{u}} \, \frac{du}{3} \\ \\ \int\limits^2_{-2} {u^{ \frac{1}{2} }} \, du \\ \\ \frac{u^{ \frac{1}{2} +1}}{ \frac{1}{2}+1 } \ |^2_{-2} \\ \\ \frac{u^{ \frac{3}{2}}}{ \frac{3}{2} } \ |^2_{-2} \\ \\ \frac{2{ \sqrt{(3x+7)^3}}}{ 3} \ |^2_{-2} \\ \\ \frac{2}{3}[ \sqrt{(3.2+7)^3}- \sqrt{(3(-2)+7)^3}] \\ \\ \frac{2}{3}[ \sqrt{13^3}- \sqrt{1^3}]$

$\frac{2}{3}[ \sqrt{13^2.13}- 1] \\ \\ \boxed{\frac{2}{3}[ 13\sqrt{13}- 1] }$

Answer 2

∫3√(3x + 7)dx  entre -2 e 2
u = 3x + 7 ⇒ du = 3dx  ⇒ dx = du/3
3∫√(u)du/3 = 3/3∫u^1/2du = [u^3/2]/3/2  = 2/3[(3x + 7)^3/2] entre -2 e 2
2/3[(3×2 + 7)^3/2 - (3× -2 + 7)^3/2]
2/3[13√13 - 1]
Resposta: 4ª alternativa