Exercise 5.  [Graphics:Images/FunTheoremCalculusModHome_gr_50.gif],  where C is the line segment  [Graphics:Images/FunTheoremCalculusModHome_gr_51.gif].  

Solution 5.

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/FunTheoremCalculusModHome_gr_52.gif].  

Solution.  The function  [Graphics:../Images/FunTheoremCalculusModHome_gr_53.gif]  is analytic on the entire complex plane, except at the origin,

and   [Graphics:../Images/FunTheoremCalculusModHome_gr_54.gif]  is analytic in the half plane [Graphics:../Images/FunTheoremCalculusModHome_gr_55.gif] that contains the  line segment joining  [Graphics:../Images/FunTheoremCalculusModHome_gr_56.gif],  

and   [Graphics:../Images/FunTheoremCalculusModHome_gr_57.gif]  has  [Graphics:../Images/FunTheoremCalculusModHome_gr_58.gif]  is an antiderivative.

Thus,  we can use Theorem 6.9 to obtain  

                              [Graphics:../Images/FunTheoremCalculusModHome_gr_59.gif]
                              

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_60.gif]

                    The path of integration is the line segment from  [Graphics:../Images/FunTheoremCalculusModHome_gr_61.gif].  

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FunTheoremCalculusModHome_gr_62.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_63.gif]



[Graphics:../Images/FunTheoremCalculusModHome_gr_64.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_65.gif]




















This solution is complements of the authors.



































 

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