Double integrals: #int_0^1 int_(y-1)^(y-3) (x^2+2y) dxdy# ?

1 Answer
Steve M
May 17, 2018

# int_0^1 \ int_(y-1)^(y-3) \ (x^2+2y) \ dx\ dy = -22/3 #

Explanation:

We seek:

# I = int_0^1 \ int_(y-1)^(y-3) \ (x^2+2y) \ dx\ dy #

Focusing first on the inner integral, and treating #y# as a constant we have:

# I = int_(y=0)^(y=1) \ int_(x=y-1)^(x=y-3) \ (x^2+2y) \ dx\ dy #
# \ \ = int_0^1 \ [ \ x^3/3+2xy \ ]_(x=y-1)^(x=y-3) \ dy #

# \ \ = int_0^1 \ ((y-3)^3/3+2(y-3)y) - ((y-1)^3/3+2(y-1)y) \ dy #

# \ \ = int_0^1 \ (y-3)^3/3+2y^2-6y - (y-1)^3/3 - 2y^2+2y \ dy #

# \ \ = int_(y=0)^(y=1) \ (y-3)^3/3-4y - (y-1)^3/3 \ dy #

# \ \ = [(y-3)^4/12 -2y^2 + (y-1)^4/12]_(y=0)^(y=1) #

# \ \ = ((-2)^4/12 -2*1 - 0) - ((-3)^4/12 - 0 - (-1)^4/12) #

# \ \ = (16/12 -2 ) - (81/12 - 1/12) #

# \ \ = 16/12 -2-81/12+1/12 #

# \ \ = -22/3 #

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