# Answer to Question #5416 in Calculus for Samantha

Question #5416

I need to find the double integral... I'll write it as it appears in my book, with the lower limit first

1.) Integral 1,4; Integral 0,y for the function (sqrt(x)) dx dy

2.) Integral -2,2; Integral -2,2 for the function (x+y) dy dx

1.) Integral 1,4; Integral 0,y for the function (sqrt(x)) dx dy

2.) Integral -2,2; Integral -2,2 for the function (x+y) dy dx

Expert's answer

1.) Integral 1,4; Integral 0,y for the function (sqrt(x)) dx dy

Integral 0,y for (sqrt(x)) =

& = Integral 0,y for (x^0.5) =

& = (x^1.5)/1.5 from x=0 to x=y =

& = (y^1.5)/1.5 - 0 =

& = (y^1.5)/1.5

Integral 1,4 for (y^1.5)/1.5 dy =

& = (y^2.5)/(2.5*1.5) from y=1 to y=4 =

& = (y^2.5)/3.75 from y=1 to y=4 =

& = (4^2.5)/3.75 - (1^2.5)/3.75 =

& = 32/3.75 - 1/3.75 = 31/3.75 = 124/15.

2.) Integral -2,2; Integral -2,2 for the function (x+y) dy dx

Integral -2,2 for (x+y) dy =

& = xy + y²/2 from y=-2 to y=2 =

& = (2x + 2²/2) - (-2x + (-2)²/2) =

& = 2x + 2 + 2x - 2 = 4x

Integral -2,2 for 4x dx =

& = 4x²/2 from x=-2 to x=2 =

& = 2*2² - 2*(-2)² = 8 - 8 = 0.

Integral 0,y for (sqrt(x)) =

& = Integral 0,y for (x^0.5) =

& = (x^1.5)/1.5 from x=0 to x=y =

& = (y^1.5)/1.5 - 0 =

& = (y^1.5)/1.5

Integral 1,4 for (y^1.5)/1.5 dy =

& = (y^2.5)/(2.5*1.5) from y=1 to y=4 =

& = (y^2.5)/3.75 from y=1 to y=4 =

& = (4^2.5)/3.75 - (1^2.5)/3.75 =

& = 32/3.75 - 1/3.75 = 31/3.75 = 124/15.

2.) Integral -2,2; Integral -2,2 for the function (x+y) dy dx

Integral -2,2 for (x+y) dy =

& = xy + y²/2 from y=-2 to y=2 =

& = (2x + 2²/2) - (-2x + (-2)²/2) =

& = 2x + 2 + 2x - 2 = 4x

Integral -2,2 for 4x dx =

& = 4x²/2 from x=-2 to x=2 =

& = 2*2² - 2*(-2)² = 8 - 8 = 0.

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