Question

CALCULES A INTEGRAL DUPLA

Answer 1

Resposta:

Explicação passo-a-passo:

$\int _{0}^{3} \int_{0}^{1} (6x^2y^3-5y^4)dydx = \int _{0}^{3}(6x^2\frac{y^4}{4}-5\dfrac{y^5}{5}) ]_{0}^{1} dx = \int _{0}^{3} (\frac{3}{2}x^2y^4- y^5)]_{0}^{1} dx = \\\int _{0}^{3} (\frac{3}{2}x^2-1) dx = \frac{3}{2}\frac{x^3}{3} -x ]_{0}^{3} = \frac{x^3}{2}-x]^{3}_{0} = \frac{3^3}{2}-3 = \frac{27}{2}-3= \frac{21}{2}$