859141 2 b 2 3x 2 0 dx 30 x3 120 dy x2 3y 2

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Unformatted text preview: = 120. dy (x2 + 3y 2 ) = dx 0 0 14.2 Evaluate the integrals by reversing the order of integration. (a) 1 1 I= dy dx √ 0 1 1 = 3 dt 0 2+ = 0 1 dx x2 2 + x3 = dy dx y √ x2 1 x3 0 2 + x3 0 √ 42 2 2+t = √ − ≈ 0.526161. 9 3 (b) 2 I= dx y sin(x2 ) = dy = = 1 2 dx sin(x2 ) y2 0 4 dx x sin(x2 ) = 0 2 √ 4 4 dy y = 0 0 1 4 16 0 4 x dt sin(t) = − 1 cos(x2 ) 4 0 x=4 y2 dx sin(x2 ) 2 = x=0 14.3 The average is x+1 4 2 1 dy 6x2 y = dx area 1 9 2 897 2 8073 = = 179.4 . =· 9 10 5 4 1 y =0 1 (1 − cos 16) 4 sin 8 ≈ 0.489415 . 2 Ave = √ y= x dx 3x4 + 6x3 − 9x2...
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This document was uploaded on 03/01/2014 for the course MTH 243 at Rhode Island.

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