Answer to Question #5416 in Calculus for Samantha
2011-12-01T09:32:59-05:00
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
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2011-12-01T11:06:46-0500
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.
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