Workshop_Solutions_5

0 making the substitution x xy y y y x x dx

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Unformatted text preview: dx. a 0 a f (x) dx = −a f (x) dx + −a f (x) dx. 0 Making the substitution x = x(y ) = −y ⇔ y = y (x) = −x, dx = −1 dy ⇒ dx = −dy in the first integral on the right-hand side, with the new boundaries y (0) = 0 and y (−a) = −(−a) = a, gives that 0 0 f (x) dx = −a (−1) f (−y ) dy a 0 =− f (y ) dy a (since f is even, that is, f (−y ) = f (y )) a = f (y ) dy (swapping the boundaries) f (x) dx (replacing y by x). 0 a = 0 Thus we find a 0 f (x) dx = −a as required. a f (x) dx + −a a f (x) dx = 0 a f (x) dx + 0 a f (x) dx = 2 0 f (x) dx, 0 2 3 Mathematical Methods for Physics 1, Solutions to Workshop Sheet 5 3. (a) By making the substitution y = 2 t or otherwise, evaluate the indefinite integral I= cos(2t) dt. (b) Using the addition theorem for cos(t + s), show that cos(2t) = 1 − 2 sin(t) 2 = 2 cos(t) 2 − 1. (c) Use (a) and (b) to evaluat...
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This document was uploaded on 01/31/2014.

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