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Example 1475 Evaluating an Integral Using the change of variables u = x − y and v = x y, evaluate the integral ∬R(x − y)ex2 − y2dA, where R is the region bounded by the lines x y = 1 and x y = 3 and the curves x2 − y2 = − 1 and x2 −And L(cu) = (cu) t (cu) xx= cu t cu xx= c(u t u xx) = cLu So, indeed, (16) is a linear equation, since it is given by a linear operator To understand how linearity can fail, let us see what goes wrong forL(x;u;v) = g(u;v) In other words, g(u;v) is a lower bound on f?for any uand v 0 Note that g(u;v) = (bTu hTv if c= ATu GTv 1 otherwise This second explanation reproduces the same dual, but is actually completely generaland applies to arbitrary optimization problems 3 1 ƒVƒ‡[ƒgƒ{ƒu ŠÛ Šç Ž—‡‚¤ "¯Œ^