Exercise 2.  [Graphics:Images/FunTheoremCalculusModHome_gr_12.gif],  where C is the line segment from  [Graphics:Images/FunTheoremCalculusModHome_gr_13.gif].

Solution 2.

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/FunTheoremCalculusModHome_gr_14.gif][Graphics:../Images/FunTheoremCalculusModHome_gr_15.gif].  

Solution.  The function  [Graphics:../Images/FunTheoremCalculusModHome_gr_16.gif]  is analytic on the entire complex plane, which is simply connected,

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_17.gif]  has  [Graphics:../Images/FunTheoremCalculusModHome_gr_18.gif]  as an antiderivative.

Thus,  we can use Theorem 6.9 to obtain  

                              [Graphics:../Images/FunTheoremCalculusModHome_gr_19.gif]    
                              

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_20.gif]

                    The path of integration is the line segment from  [Graphics:../Images/FunTheoremCalculusModHome_gr_21.gif].

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FunTheoremCalculusModHome_gr_22.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_23.gif]



[Graphics:../Images/FunTheoremCalculusModHome_gr_24.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_25.gif]




















This solution is complements of the authors.



































 

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