Exercise 8.  [Graphics:Images/FunTheoremCalculusModHome_gr_93.gif],  where C is the line segment from  [Graphics:Images/FunTheoremCalculusModHome_gr_94.gif].  

Solution 8.

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/FunTheoremCalculusModHome_gr_95.gif].  

Solution.  The function  [Graphics:../Images/FunTheoremCalculusModHome_gr_96.gif]  is analytic on the entire complex plane, which is simply connected,

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_97.gif]  has  [Graphics:../Images/FunTheoremCalculusModHome_gr_98.gif]  as an antiderivative.  

Thus,  we can use Theorem 6.9 to obtain  

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_99.gif]  
                    

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_100.gif]

                    The path of integration is the line segment from  [Graphics:../Images/FunTheoremCalculusModHome_gr_101.gif].  

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FunTheoremCalculusModHome_gr_102.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_103.gif]



[Graphics:../Images/FunTheoremCalculusModHome_gr_104.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_105.gif]




















This solution is complements of the authors.



































 

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