MATH2010-assign4 - Math-2010 Problem Set 4 Hyeon Kang 1 z =...

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Math-2010 Problem Set 4 Hyeon Kang 1 (a) z = e - ( x 2 ) V = R b a 2 πxf ( x ) dx . V = lim b →∞ R b 0 2 πxe - x 2 dx = π (lim b →∞ ( e - x 2 ) b 0 ) = π (b) p. 979 # 37 (b) R 0 xe - x dx. Let u = x. u 2 = x and dx = 2 u du. = lim b →∞ R b 0 u e - u 2 2 u du = 2( R 0 u 2 e - u 2 du ) . Integrate by parts. .. j = u , dk = ue - x 2 du. dj = du k = - e - x 2 2 . lim b →∞ R 0 u 2 e - u 2 du = lim b →∞ ( xe - x 2 2 | b 0 + R b 0 e - x 2 2 dx. ) = 0 + (1/2) R 0 e - x 2 dx = 1 2 * ((1 / 2) R -∞ e - x 2 dx = π 4 . ( Even function can be divided.) Then 2( R 0 u 2 e - u 2 du ) = π 2 2 p. 1032 # 11, p. 1033 # 29 #11) Plot II . Examine that since F ( x, y ) = < y, x > , the vector field must be perpendic- Also, the first quadrant has both positive x and y components, second quad- rant has positive x components and negative y components, etc. Plot II is the only one that fits this description. #29) graph
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This note was uploaded on 03/27/2008 for the course MATH 2010 taught by Professor Herron during the Spring '08 term at Rensselaer Polytechnic Institute.

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MATH2010-assign4 - Math-2010 Problem Set 4 Hyeon Kang 1 z =...

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