Exe M1-17 - Exercises module 1 mathematics PDF

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mathematics for economists M1 2017/18: exercise session 17

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improper integrals Exercise 241. Explain what is an improper integral. answer: an improper integral is a definite integral that has either one or both limits of integration infinite or an integrand that tends to plus (or minus) infinity at one or more points in the range of integration.

Exercise 242. Explain how it is possible to decide whether the improper integrals

1.

Z

+1

f (x) dx

2.

a

Z

a

f (x) dx

3.

1

Z

+1

f (x) dx 1

with a ∈ R and f (x) a continuous function in the interval of integration, exist.

answer: 1. The integral is computed as the limit for h approaching plus infinity of the integral obtained by replacing the infinite Z +1 Z h f (x) dx . The existence of this limit corresponds to the existence of the f (x) dx = lim

limit of integration with h,

a

h!+1

a

improper integral.

2. The integral is computed as the limit for h approaching minus infinity of the integral obtained by replacing Z a Z a the infinite limit of integration with h, f (x) dx = lim f (x) dx . Again, the existence of this limit corresponds to the h!1 h 1 Z +1 Z +1 Z a f (x) dx + existence of the improper i ntegral. 3. The integral is rewritten as f (x) dx with a any real f (x) dx = a

1

1

number. The improper integral exists if both parts exi st in the sense given above.

improper integrals with infinite limits of integration Exercise 243. Does the improper integral

Z

1 1

1 dx x

exist? Explain solution: the integral does not exist because the limit

lim

h!+1

ln h ! +1 as h ! +1.

Z h 1

x

1

dx =

lim

h!+1

⇣

h

ln x| 1

⌘

=

lim ln h does not exit because

h!+1

Exercise 244. Does the improper integral

Z

1 1

1 dx x2

exist? Explain solution: the integral exists because the limit

lim

h!+1

Z h 1

1 x2

dx =

68

lim

h!+1



◆ ✓  ! 1 1 h  + 1 = 1. = lim  h!+1 x 1 h

Exercise 245. Does the improper integral

Z

1

1 √ dx x

1

exist? Explain solution: the integral does not exist because the limit because

p

lim

h!1

h ! +1 as h ! +1.

⇣ p ⌘ ⌘ ⇣ p 1 h lim 2 x| 1 = lim 2 h  2 does not exit p dx = h!+1 h!+1 x

Z h 1

Exercise 246. Find all α > 0 such that

Z

1 dx xα

1 1

exists. solution: α > 1.

Exercise 247. Does the improper integral

Z

0 1

x2

x dx +1

exist? Explain. Z 0

solution: the integral does not exist because the limit lim h!1 ⇣ ⌘ does not exist as ln h2 + 1 ! +1 as h ! 1.

h

x

dx =

x2 + 1

lim

h!1

✓

1 2

⌘ ⇣ ⇣ ⌘0 ◆ 1 2 2 ln h + 1 ln x + 1  = lim h!1 2 h

Exercise 248. The function f is defined for x > 0 by f (x) =

convergence of Z 1 f (x)dx.

ln x . Examine the x3

1

solution:

Z 1

f (x)dx is convergent because

1

l’Hopital’s rule

Z 1 ln x x3

1

lim

h!+1

ln h h2

=

lim

h!+1

1 h

2h

=

lim

1

h!+1 2h2

dx =

Z h ln x

lim h!+1

1

x3

dx =

lim h!+1



= 0.

 ! 1 1 + 2 ln x  h = , in fact by  4 4x2 1

Exercise 249. Does the improper integral

Z

+1

xex dx 0

exist? Explain. solution: the integral exists because the limit exists; in fact by l’Hopital’s rule

lim h!+1

lim

h!+1 h

(1 + h)e

=

Z h

x

xe

0

lim

h!+1

dx =

1+h eh

69

lim

h!+1

=

lim

✓

1

h!+1 eh

= 0.

 ◆

x  0  h

(1 + x)e

=

lim

h!+1

⇣

h

1  (1 + h)e

⌘

= 1

review exercises miscellaneous exercises Exercise 250. Let f (x) =

√

1 − x.

a) Specify the domain of f . b) Is f a monotone function? Explain. c) Is f a concave function? Explain d) Find the quadratic approximation of f about x0 = 0. f (x) . x!1 x − 1

e) Find lim

f ) Write down the equation of the tangent line at x0 = −8. solution: a) The square root is defined if and only if 1  x  0, hence D = {x 2 R : x  1}. 0

f (x) =

 p1 2 1x

is everywhere negative on D; the function f is monotone decreasing.

f 00 (x) =  q 1 4

(1x)3

is evrywhere negative on D.

b) Yes, because the derivative

c) Yes, because the second derivative

d) The quadratic approximation about x0 of a twice differentiable function

f is given by f (x) ' f (x0 ) + f 0 (x0 )(x  x0 ) + 21f 00 (x0 )(x  x0 )2 ; since f (0) = 1, f 0 (0) = 21 and f 00 (0) =  41 we obtain x2  1 x + 1. f (x) '  1 8 2

1 ; since numerator and denominator are differential e) The limit yields the indeterminate form 1 p

1x

functions of x, by l’Hˆ opital’s rules lim x!1 x1

= lim x!1



2

p1 1x 1

= limx!1  p 1 2

1x

= 0.

f) The equation of

0

the tangent line at x0 the the graph of the differentiable function f is given by y = f (x0 ) + f (x0 )(x  x0 ); Since f (8) = 3 and 5. f 0 (8) =  61 we obtain y =  61 x + 3

Exercise 251. Find the limit

lim x2 ln x2 .

x!0+

Explain. solution: the limit yields the indeterminate form 0 · (1). After rewriting x2 ln x2 as the quotient of two differentiable functions 2

of x, l’Hˆ opital’s rules yield lim x!0+ x2 ln x2 = limx!0+ 2 ln x = lim x!0+ x2 1  3 2 x

= limx!0+ (x2 ) = 0.

x

Exercise 252. Find all x ∈ R such that 1 X (2x − 1)n 0

is convergent. solution: the geometric series is convergent if and only if |2x  1| < 1, that is for x 2 (0, 1).

70

Exercise 253. Does the equation

x4 + 1 = 3x have a solution in the interval [0, 1]? Explain. solution: the solutions of the equation x4 + 1 = 3x correspond to the zeros of the continuous function f (x) = x4  3x + 1. Since f (0) = 1 and f (1) = 1 the intermediate value theorem guarantees that the function f has at least one zero in the interval (0, 1). Hence the equation at least one solution in the interval.

Exercise 254. Find a function F such that F 0 (x) = 3x2 − 6x and F (1) = 2.

Explain.

solution: the function F is an indefinite integral of 3x2  6x, hence F (x) = 3

2

imposing 2 = F (1) = 1  3 + C we obtain C = 4. Hence, F (x) = x  3x + 4.

R

3x2  6x dx = x3  3x2 + C, with C 2 R. By

Exercise 255. Write down

x3 − 3x2 + 2x − 1 x2 + 1 as a sum of a polynomial and a proper rational function. solution:

x3 3x2 +2x1 = x2 +1

x  3 + x+2 . 2 x +1

Exercise 256. Consider

Z

1

x3 dx. 4 1 1 + x

Is this a proper integral? Explain. Evaluate the integral. solution: yes, it is a proper integral because both limits of integration are finite and the integrand is bounded in the interval of R1 x3 dx = integration. By setting t = x4 + 1, hence dt = 4x3 dx or x3 dx = 41 dt and since for x = ±1, t = 2, we obtain 1 1+x4 1 R2 1 4 2 t dt = 0 because the limits of integration are identical.

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