Question

o valor da integral ∫03∫02(4-y2) é:

Answer 1

Olá!

$\displaystyle\int_0^3\int_0^24-y^2dydx=\int_0^3\left(4y-\dfrac{y^3}{3}\right)\bigg{|}_0^2\;dx =\\ \\ \\ = \int_0^3\left[\left(8-\dfrac{8}{3}-0\right)\right]dx=\dfrac{16}{3}\int_0^3dx=\dfrac{16}{3}x\bigg{|}_0^3=\dfrac{16}{3}\cdot (3-0)=\\ \\ \\ = \dfrac{16}{3}\cdot 3=16.$

Bons estudos!