Exercise 1.  [Graphics:Images/FunTheoremCalculusModHome_gr_1.gif],  where C is the line segment from  [Graphics:Images/FunTheoremCalculusModHome_gr_2.gif].   

Solution 1.

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/FunTheoremCalculusModHome_gr_3.gif].

Solution.  The function  [Graphics:../Images/FunTheoremCalculusModHome_gr_4.gif]  is analytic on the entire complex plane, which is simply connected,

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_5.gif]  has  [Graphics:../Images/FunTheoremCalculusModHome_gr_6.gif]  as an antiderivative.

Thus,  we can use Theorem 6.9 to obtain  

                              [Graphics:../Images/FunTheoremCalculusModHome_gr_7.gif]    
                              

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_8.gif]

                    The path of integration is the line segment from  [Graphics:../Images/FunTheoremCalculusModHome_gr_9.gif].  

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FunTheoremCalculusModHome_gr_10.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_11.gif]



















This solution is complements of the authors.



































 

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