Calcule ∬_R^() ( x + y ) d A , onde R é a região no (a) Exercício 10 Passo 1Fala galera! No exercício 10 temos que: int_0^1 int_(x^2)^(sqrt(x)) f (x , y) d y d x int_0^1 int_(y^2)^(sqrt(y)) f (x , y) d x d y Passo 2Vamos resolver a integral dubla. int_0^1 int_(x^2)^(sqrt(x)) f (x , y) d y d x = int_0^1 int_(x^2)^(sqrt(x)) ( x + y ) d y d x = int_0^1 [x y + (y^2)/(2)]_(x^2)^(sqrt(x)) d x = int_0^1 (x sqrt(x) + (( sqrt(x) )^2)/(2) - x x^2 - (( x^2 )^2)/(2)) d x = int_0^1 (x^(3 // 2) + (x)/(2) - x^3 - (x^4)/(2)) d x = [(2 x^(5 // 2))/(5) + (x^2)/(4) - (x^4)/(4) - (x^5)/(10)]_0^1 = (2 * 1^(5 // 2))/(5) + (1^2)/(4) - (1^4)/(4) - (1^5)/(10) = (2)/(5) - (1)/(10) = (3)/(10) int_0^1 int_(y^2)^(sqrt(y)) f (x , y) d x d y = int_0^1 int_(y^2)^(sqrt(y)) (x + y) d x d y = int_0^1 [(x^2)/(2) + x y]_(y^2)^(sqrt(y)) d y = int_0^1 ((( sqrt(y) )^2)/(2) + sqrt(y) y - (( y^2 )^2)/(2) - y^2 y) d y = int_0^1 ((y)/(2) + y^(3 // 2) - (y^4)/(2) - y^3) d y = [(x^2)/(4) + (2 x^(5 // 2))/(5) - (x^5)/(10) - (x^4)/(4)]_0^1 = (1^2)/(4) + (2 * 1^((5)/(2)))/(5) - (1^5)/(10) - (1^4)/(4) = (2)/(5) - (1)/(10) = (3)/(10) Resposta (3)/(10) MOSTRAR SOLUÇÃO COMPLETA