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# 15.2 - 15.2 ITERATED INTEGRALS KIAM HEONG KWA 1 Iterated...

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Unformatted text preview: 15.2: ITERATED INTEGRALS KIAM HEONG KWA 1. Iterated Integrals Let f ( x, y ) be a continuous function on the rectangle R = [ a, b ] × [ c, d ]. By holding x fixed, we obtain a function g ( y ; x ) of y , parametrized by x , with the property that g ( y ; x ) = f ( x, y ) ( x is fixed) for all y ∈ [ c, d ]. In particular, for each x ∈ [ a, b ], we obtain an integral integraltext d c g ( y ; x ) dy that depends on x . Therefore by setting A ( x ) = integraldisplay d c g ( y ; x ) dy for each x ∈ [ a, b ], we have A as a function of x on [ a, b ]. It is customary to write A ( x ) = integraldisplay d c f ( x, y ) dy with the tacit understanding that the integration is done with re- spect to y while x is held fixed. This procedure is referred to as the partial integration with respect to y . Remark 1. In the case f ( x, y ) ≥ , A ( x ) can be interpreted as the area of a cross-section of the solid lying above the xy-plane and below the graph of f . The cross-section is labeled by the x-coordinate. For different values of x , one gets different cross-sections as well as different cross-sectional areas. In fact, A is a continuous function since so is f . Hence we can integrate A with respect to x and obtain integraldisplay b a A ( x ) dx = integraldisplay b a bracketleftbiggintegraldisplay d c f ( x, y ) dy bracketrightbigg dx, Date : October 10, 2010. 1 2 KIAM HEONG KWA an iterated integral . Note the order of integration. Very often we omit the brackets and write (1.1) integraldisplay b a integraldisplay d c f ( x, y ) dy dx = integraldisplay b a bracketleftbiggintegraldisplay d c f ( x, y ) dy bracketrightbigg dx. Likewise, if we integrate with respect to x first, then we obtain the iterated integral (1.2) integraldisplay d c integraldisplay b a f ( x, y ) dx dy = integraldisplay d c bracketleftbiggintegraldisplay b a f ( x, y ) dx bracketrightbigg dy. Example 1 (Problem 4 in the text) . Calculate integraltext 1 integraltext 2 1 (4 x 3- 9 x 2 y 2 ) dy dx . Solution. For each x ∈ [0 , 1] , one has integraldisplay 2 1 (4 x 3- 9 x 2 y 2 ) dy = bracketleftbigg 4 x 3 y- 9 x 2 · y 3 3 bracketrightbigg y =2 y =1 = bracketleftbig 4 x 3 y- 3 x 2 y 3 bracketrightbig y =2 y =1 = (4 x 3 · 2- 3 x 2 · 2 3 )- (4 x 3 · 1- 3 x 2 · 1 3 ) = 8 x 3-...
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15.2 - 15.2 ITERATED INTEGRALS KIAM HEONG KWA 1 Iterated...

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