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86 Chap- 7 b <br />A, solution sattisfy1mg the differeztW equation and the boundary conditions <br />is? <br />® 2 <br />x e U du <br />Amt an 4 <br />( x <br />'k <br />Values of this integral nay be obtainer from 'T'able 9. At x 0 this becomes : <br />+E <br />??o = 2m= (7-2) <br />It will be of interest to co2pute the flaw f passing between planes a unit <br />distance apart. 'his flaw is, to a first approximation: <br />f--KD ? -- /1 ? 2 x 2w e?+ a du <br />t <br />u <br />x le, <br />'74=a t? <br />To obtain this result the procedure for differentiating an integral has been <br />folloved. Mds was described previously. In the present case, however, the <br />vriation is ion the lower licit which introduces a negative sign. An evalua- <br />tion of the inttegaal is needed. To obtain this integrate by parts with <br />2 <br />u <br />ul i e dr l ` du <br />U. <br />2 <br />dual - -toe -U do vi = I <br />ua <br />Mena since <br />h <br />l <br />G <br />1"L " olvl - jvldl