Workshop_Solutions_5

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 01/31/2014.

Ask a homework question - tutors are online