Exercise 13.  Evaluate  [Graphics:Images/ContourIntegralModHome_gr_604.gif],  where C is the straight-line segment joining  [Graphics:Images/ContourIntegralModHome_gr_605.gif].

Solution 13.

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/ContourIntegralModHome_gr_606.gif].  

                    [Graphics:../Images/ContourIntegralModHome_gr_607.gif]

                    The curve  [Graphics:../Images/ContourIntegralModHome_gr_608.gif],  for  [Graphics:../Images/ContourIntegralModHome_gr_609.gif].

Solution.  The function is  [Graphics:../Images/ContourIntegralModHome_gr_610.gif]  and the curve is  [Graphics:../Images/ContourIntegralModHome_gr_611.gif]   for  [Graphics:../Images/ContourIntegralModHome_gr_612.gif]  and we obtain

                     [Graphics:../Images/ContourIntegralModHome_gr_613.gif]   and   [Graphics:../Images/ContourIntegralModHome_gr_614.gif],  
                     
                    then

                     [Graphics:../Images/ContourIntegralModHome_gr_615.gif]   

The last two real integrals are computed using

                    [Graphics:../Images/ContourIntegralModHome_gr_616.gif]  
                    
                    
                    and
                    

                    [Graphics:../Images/ContourIntegralModHome_gr_617.gif]  

We are done.  

Aside.  After we have developed the Cauchy-Goursat Theorem in Section 6.3 and the Fundamental Theorem of Calculus in Section 6.4

we will be able to compute the integral in an easy fashion:

                    [Graphics:../Images/ContourIntegralModHome_gr_618.gif]   

We are really done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ContourIntegralModHome_gr_619.gif]

[Graphics:../Images/ContourIntegralModHome_gr_620.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2008 John H. Mathews, Russell W. Howell