Evaluate
"\\int\\int_D(3-x-y)dA"
[dA means dxdy or dydx]
where D is the triangle in the (x, y) plane bounded by the x-axis and the lines y = x
and x = 1.
Solution.
"\\iint_D(3-x-y)dxdy=\\int^1_0(\\int^{y=x}_{y=0}(3-x-y)dy)dx=\\int^1_0((3y-xy-\\dfrac{y^2}{2})\\mid^{y=x}_{y=0})dx="
"=\\int^1_0(3x-x^2-\\dfrac{x^2}{2})dx=(3\\dfrac{x^2}{2}-\\dfrac{x^3}{3}-\\dfrac{x^3}{6})\\mid^1_0=\\dfrac{3}{2}-\\dfrac{1}{3}-\\dfrac{1}{6}=1."
Answer: 1.
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