Exercise 9.  [Graphics:Images/FunTheoremCalculusModHome_gr_106.gif],  where C is the line segment from  [Graphics:Images/FunTheoremCalculusModHome_gr_107.gif].  

Solution 9.

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/FunTheoremCalculusModHome_gr_108.gif][Graphics:../Images/FunTheoremCalculusModHome_gr_109.gif].  

Solution.  The function  [Graphics:../Images/FunTheoremCalculusModHome_gr_110.gif]  is analytic on the entire complex plane, which is simply connected,

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_111.gif]  has  [Graphics:../Images/FunTheoremCalculusModHome_gr_112.gif]  as an antiderivative.  

Thus,  we can use Theorem 6.9 to obtain  

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_113.gif]  
                    

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_114.gif]

                    The path of integration is the line segment from  [Graphics:../Images/FunTheoremCalculusModHome_gr_115.gif].  

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FunTheoremCalculusModHome_gr_116.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_117.gif]




















This solution is complements of the authors.



































 

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