ECON3120/4120 Mathematics 2, spring 2009

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1 University of Oslo Department of Economics Arne Strøm ECON3/4 Mathematics, spring 9 Problem soltions for Seminar 4, 6 Febrary 9 (For practical reasons some of the soltions may inclde problem parts that are not on the problem list for this seminar.) EMEA, 9.7. (= MA I,.9.) (a) d = lim 3 b (b) The integral b 3 d = lim b b ( = lim b b + ) =. b b d = / d = b = b does not converge to any limit as b. Hence, the integral (c) e d = lim e d = a a y lim a a e = lim ( a ea )=. d diverges. y = a a ε a Problem 9.7.(d) (d) If we introdce = a as a new variable we get d = a d = d = + C = a + C. d and a matv9sem4svar

2 Therefore, a a ε d = lim a ε + d = lim a ε + a ε a = lim ε + ( a (a ε) + a ) = a = a (since a>). Note that we let ε tend to from the right so that a ε tends to a from the left. EMEA, (= MA I,.9.7) The integrand, f() = + + 3, is defined only in the open interval (, 3) and tends to at both ends of this interval. In order to show that the integral converges, we se the recipe in formla (9.7.3) on page 3 (formla (.9.3) on page 36 in MA I): We split the interval at an arbitrary point, at =, for instance, and then show that the integrals over (, ] and [, 3) both converge. The indefinite integral is f() d = F ()+C, where F () = + 3, and it is clear that and 3 f() d = lim a +(F () F (a)) = F () F ( ) f() d = lim b 3 (F (b) F ()) = F (3) F () eist. It follows that f is integrable over (, 3) and 3 f() d = Eam problem f() d + 3 f() d = F (3) F ( )= 5+ 5=4 5. (a) Using the rles (a + b) n d = (a + b)n+ a(n +) + C (a, n ), and we get e a = ea a + C (a ), (( ) + e ) d = ( )3 3 + e + C. matv9sem4svar

3 (b) Here we cold se polynomial division to simplify the fraction, bt it is a little easier to se the sbstittion =. Then = +,d = d, and ( +) d = ( +) d = d ( ) ( ) = d = ln + C = ln + C. (This can also be written as ln + C, with C = C +.) (c) The innermost integral is = ( + y) d = = + y = y + + y +, and the doble integral is therefore ( y + + ) dy = ( ln(y +)+ln(y + )) y + =( ln3+ln) ( ln + ln ) = ln ln 3 = ln(4/3). Eam problem 7 (a) The domain bonded by the crve y =4 /( + ), the -ais, and the straight line = 4 is shaded in the figre below. y y = Problem 7 4 The area of this domain is A = + d. With the sbstittion =+ we get =( ), d =( ) d, and A = = =4 = 4 4( ) ( ) d = 4 8( ) d = 4 ( (4 3 +3ln) = (64 8+3ln4) (6 64+3ln) =3ln 6 ( 6.87). 3 ) d matv9sem4svar

4 (b) L Hôpital s rle yields a a lim a a = a ln a a a = lim = a a (ln a ). a Eam problem 77 (i) We first calclate the indefinite integral. Integration by parts gives ( + ) /3 d = 3 4 ( + )4/3 3 ( + ) 4/3 d 4 = 3 4 ( + )4/3 9 8 ( + )7/3 + C The definite integral is then 6 6 ( 3 ( + ) /3 d = 4 ( + )4/3 9 ) ( + )7/3 8 = 9 84/3 9 ( 8 87/ ) = , where we have sed that 8 /3 = 3 8=. Alternatively, we can se sbstittion and calclate as follows: Introdce =(+) /3 as a new variable. That gives = 3, d =3 d, and ( + ) /3 d = ( 3 )3 d = ( ) d = C = 3 7 ( + )7/3 3 ( + )4/3 + C. (This is indeed eqal to the indefinite integral we fond above, althogh it does not look that way at first glance.) We the calclate the definite integral as before. However, we can also se formla () on page 333 in EMEA (page 355 in MA I). That will give s 6 ( 3 ( + ) /3 = ( ) d = ) 4 etc. (ii) Here we se the sbstittion z = 3 = /3, which gives = z 3 and d =3z dz. The integral then becomes e 3 d = e z 3z dz =3 z e z dz. In order to find the last integral, we se integration by parts twice: z e z dz = z e z ze z dz = z e z ( ze z e z dz ) = z e z ze z + e z dz = z e z ze z +e z + C. 4 matv9sem4svar

5 Then where C =3C. e 3 d =3(z e z ze z +e z + C) =(3 /3 6 /3 +6)e 3 + C, Eam problem 97 (a) We have ϕ()=ln ln = ln. Frther, ϕ () = + + = ( + )( +) > for all, so ϕ is strictly increasing. Finally, ϕ() =ln + + =ln+/ as. +/ It follows that the range (Norwegian: verdimengden ) of ϕ is V ϕ =[ ln, ). Note that it is not enogh to observe that ϕ() = ln and lim ϕ() =. The fnction vale, ϕ(), might conceivably go both p and down as rns from to, so that ϕ might take on vales otside the interval [ ln, ). Bt once we know that the fnction is increasing, this cannot happen. (b) The fnction ϕ is strictly increasing throghot its domain (Norwegian: definisjonsområde ), namely the interval [, ). Therefore it has an inverse ϕ defined on the range V ϕ =[ ln, ) of ϕ. We can find a formla for the inverse in the following way: y = ϕ () ϕ(y) = ln y + y + = y + y + = e y +=e (y +) y( e )=e y = e e. (c) In order to find the inverse of the fnction ϕ, we first need to check whether ϕ really has an inverse. We know that ϕ () is defined for >. Becase ϕ () = ( +) + ( +) = ( +) ( +) ( +) ( +) = 3 ( +) ( +) < for all >, the fnction ϕ () is strictly decreasing throghot its domain, D ϕ =(, ). Since ϕ () / as and ϕ () as, the range of ϕ is V ϕ =(, /). It follows that ϕ has an inverse fnction, defined on (, /).. Let h =(ϕ ) be this inverse. Then h is defined on (, /). We have y = h() ϕ (y) = (y + )(y +) = (y + )(y +)= ( ) y +3y + =. 5 matv9sem4svar

6 It is tacitly nderstood here that y lies in V h = D ϕ =(, ) and that lies in D h =(, /). Eqation ( ) is a qadratic eqation for y. If we solve this eqation we get y = 3 ± 9 4( ) = 3 ± + 4 = 3 ± 4 +. Since y is spposed to be positive, it is clear that we have to choose the + sign in front of the sqare root. Hence, (ϕ ) () =h() = matv9sem4svar