Exercise 3.  [Graphics:Images/FunTheoremCalculusModHome_gr_26.gif],  where C is the line segment from  [Graphics:Images/FunTheoremCalculusModHome_gr_27.gif].  

Solution 3.

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/FunTheoremCalculusModHome_gr_28.gif].  

Solution.  The function  [Graphics:../Images/FunTheoremCalculusModHome_gr_29.gif]  is analytic on the entire complex plane, which is simply connected,

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_30.gif]  has  [Graphics:../Images/FunTheoremCalculusModHome_gr_31.gif]  as an antiderivative.

Thus,  we can use Theorem 6.9 to obtain  

                               [Graphics:../Images/FunTheoremCalculusModHome_gr_32.gif]   
                               

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_33.gif]

                    The path of integration is the line segment from  [Graphics:../Images/FunTheoremCalculusModHome_gr_34.gif].  

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FunTheoremCalculusModHome_gr_35.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_36.gif]




















This solution is complements of the authors.



































 

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