Why is ln x leq sqrt x



Task 1

Calculate $ \ D \ int_0 ^ \ infty \ frac {1} {\ sqrt {1 + x ^ 2} ^ 3} \, dx $ by first using the substitution $ x = \ sinh t $ and then $ u = substitute e ^ t $.
tip

$ \ D 1+ \ sinh ^ 2 t = \ cosh ^ 2 \ \ D \ cosh t = \ frac 12 \ left (e ^ t + \ frac {1} {e ^ t} \ right) $


solution

$ \ D x = \ sinh t, \ quad \ frac {dx} {dt} = \ cosh t = \ sqrt {1 + x ^ 2}, \ quad x = 0 \ LR t = 0, \ quad x \ to \ infty \ LR t \ to \ infty \ \ D \ RA \ int_0 ^ \ infty \ frac {1} {\ sqrt {1 + x ^ 2} ^ 3} \, dx = \ int_0 ^ \ infty \ frac { 1} {\ cosh ^ 3 t} \ cosh t \, dt = \ int_0 ^ \ infty \ frac {1} {\ cosh ^ 2 t} \, dt \ \ D e ^ t = u, \ quad \ frac {du} {dt} = e ^ t = u, \ quad t = 0 \ LR u = 1, \ quad t \ to \ infty \ LR u \ to \ infty \ \ D \ RA \ int_0 ^ \ infty \ frac {1} {\ cosh ^ 2 t} \, dt = \ int_1 ^ \ infty \ frac {4} {\ left (u + \ frac 1u \ right) ^ 2} \ frac 1u \, du = \ int_1 ^ \ infty \ frac {4u} {\ left (u ^ 2 + 1 \ right) ^ 2} \, du \ \ D w = u ^ 2 + 1, \ quad \ frac {dw} {du} = 2u, \ quad u = 1 \ LR w = 2, \ quad u \ to \ infty \ LR w \ to \ infty \ \ D \ RA \ int_1 ^ \ infty \ frac {4u} {\ left (u ^ 2 + 1 \ right) ^ 2} \, du = 2 \ int_2 ^ \ infty \ frac {1} {w ^ 2} \, dw = 2 \ left [- \ frac 1w \ right] _2 ^ \ infty = 1 $

exercise 2

Find an antiderivative to $ \ D f (x) = \ left (\ frac {x} {\ sqrt {1-x ^ 2}} \ right) ^ 3 $.
  1. with the substitution $ u = \ sqrt {1-x ^ 2} $.

  2. with the substitutions $ x = \ sin t $ and $ u = \ cos t $.

tip
  1. Case distinction.
  2. $ \ sin ^ 2 t + \ cos ^ 2 t = 1 $.

solution
  1. First let $ x \ ge 0 $.

    $ \ D u = \ sqrt {1-x ^ 2} \ quad \ LR \ quad x = \ sqrt {1-u ^ 2} \ quad \ mbox {and} \ quad \ frac {dx} {du} = - \ frac {u} {\ sqrt {1-u ^ 2}} \ \ D \ RA \ quad \ int \ frac {x ^ 3} {\ sqrt {1-x ^ 2} ^ 3} \, dx = - \ int \ frac {\ sqrt {1-u ^ 2} ^ 3} {u ^ 3} \, \ frac {u} {\ sqrt {1-u ^ 2}} \, du \ \ D = - \ int \ frac {1-u ^ 2} {u ^ 2} \, du = \ frac {1} {u} + u + C = \ frac {1} {\ sqrt {1-x ^ 2}} + \ sqrt {1-x ^ 2} + C = \ frac {2-x ^ 2} {\ sqrt {1-x ^ 2}} + C $

    If $ x \ le 0 $, then $ \ D x = - \ sqrt {1-u ^ 2} $ and $ \ D \ frac {dx} {du} = \ frac {u} {\ sqrt {1- u ^ 2}} $. These two changed signs cancel each other out in the second integral,
    and the antiderivative has the same form as above.

  2. Since $ f $ is only defined on $ [- 1,1] $, $ x = \ sin t $ for $ t \ in [- \ frac \ pi 2, \ frac \ pi 2] $ is a bijective mapping. The cosine is positive on this interval.
    Therefore one has

    $ \ D x = \ sin t \ quad \ RA \ quad \ sqrt {1-x ^ 2} = \ cos t \ quad \ mbox {and} \ quad \ frac {dx} {dt} = \ cos t \ \ D \ int \ frac {x ^ 3} {\ sqrt {1-x ^ 2} ^ 3} \, dx = - \ int \ frac {\ sin ^ 3 t} {\ cos ^ 3 t} \ cos t \, dt = \ int \ frac {\ sin ^ 3 t} {\ cos ^ 2 t} \, dt = \ int \ frac {1- \ cos ^ 2 t} {\ cos ^ 2 t} \ sin t \ dt $

    With $ u = \ cos t $ and $ \ D \ frac {du} {dt} = - \ sin t $ the integral becomes

    $ \ D \ int \ Big (1- \ frac {1} {u ^ 2} \ Big) \, du = u + \ frac {1} {u} + C \ \ D = \ cos (t) + \ frac {1} {\ cos (t)} + C = \ sqrt {1-x ^ 2} + \ frac {1} {\ sqrt {1-x ^ 2}} + C = \ frac {2-x ^ 2} {\ sqrt {1-x ^ 2}} + C. $

Task 3

Calculate $ \ D \ int_0 ^ 1 \ sqrt {1+ \ sqrt {x}} \, dx $
tip

Substitute the radicand.


solution

$ \ D t = 1 + \ sqrt {x} \ LR x = (t-1) ^ 2, \ quad \ frac {dx} {dt} = 2 (t-1), \ quad x = 0 \ LR t = 1, \ quad x = 1 \ LR t = 2 \ \ D \ RA \ int_0 ^ 1 \ sqrt {1+ \ sqrt {x}} \, dx = \ int_1 ^ 2 \ sqrt {t} \ cdot 2 (t-1) \, dt = 2 \ int \ left (t ^ {\ frac {3} {2}} - t ^ {\ frac {1} {2}} \ right) \, dt \ \ D = 2 \ left [\ frac {2} {5} t ^ {\ frac {5} {2}} - \ frac {2} {3} t ^ {\ frac {3} {2}} \ right] ^ 2_1 \ \ D = 4 \ left [\ frac {2} {5} 2 ^ {\ frac {5} {2}} - \ frac {2} {5} - \ frac {2} {3} 2 ^ {\ frac {3} {2}} + \ frac {2} {3} \ right] = 4 \ left [\ sqrt {2} \ left (\ frac {1} {5} \ cdot4- \ frac {1 } {3} \ cdot2 \ right) + \ left (\ frac {1} {3} - \ frac {1} {5} \ right) \ right] = \ frac {8} {15} (\ sqrt {2 } +1) $

Task 4

  1. $ \ D \ int \ frac {\ cos x} {(\ sin x + 1) \ sin x} \, dx \ quad $ Note: Substitute appropriately.

  2. $ \ D \ int x ^ 2 \ ln x \, dx $

  3. $ \ D \ int_1 ^ 3 \ frac {x ^ 2-8} {x ^ 2-16} \, dx $

tip
  1. Partial fraction decomposition.
  2. Partial integration.
  3. sent $ 0 to add $ and PBZ.

solution
  1. $ \ D t = \ sin x, \ quad \ frac {dt} {dx} = \ cos t \ LR dx = \ frac {1} {\ cos x} dt \ \ D = \ int \ frac {\ cos x} {(t + 1) t} \ frac {1} {\ cos x} \, dt = \ int \ frac {1} {(t + 1) t} \, dt \ $

    Partial fraction decomposition:

    $ \ D \ frac {1} {(t + 1) t} = \ frac At + \ frac {B} {t + 1} = \ frac {A (t + 1) + Bt} {t (t + 1) } \ \ D \ RA A (t + 1) + Bt = 1 \ LR A = 1, \ quad B + A = 0 \ LR B = -1 \ \ D = \ int \ frac 1t- \ frac { 1} {t + 1} \, dt = \ ln | t | - \ ln | t + 1 | + c \ D = \ ln | \ sin x | - \ ln | \ sin x + 1 | + c = \ ln \ left (\ frac {| sin x |} {sin x +1} \ right) + c $

  2. $ \ D \ int x ^ 2 \ ln x \, dx = \ frac 13 x ^ 3 \ ln x- \ int \ frac 13 x ^ 3 \ frac 1x \, dx \ \ D = \ frac 13 x ^ 3 \ ln x- \ frac 13 \ int x ^ 2 \, dx = \ frac 13 x ^ 3 \ ln x- \ frac 19 x ^ 3 + c \ \ D = \ frac 19 x ^ 3 (3 \ ln x -1) + c $

  3. $ \ D \ int_1 ^ 3 \ frac {x ^ 2-8} {x ^ 2-16} \, dx = \ int_1 ^ 3 \ frac {x ^ 2-16 + 8} {x ^ 2-16} \ , dx = \ int_1 ^ 3 1+ \ frac {8} {x ^ 2-16} \, dx $

    Patial fracture dissection

    $ \ D \ frac {8} {x ^ 2-16} = \ frac {A} {x-4} + \ frac {B} {x + 4} = \ frac {A (x + 4) + B ( x-4)} {x ^ 2 + 16} \ \ D \ RA A + B = 0, \ quad 4A-4B = 8 \ LR A = 1, B = -1 \ \ D = \ int_1 ^ 3 \ left (1+ \ frac {1} {x-4} - \ frac {1} {x + 4} \ right) \, dx = \ big [x + \ ln | x-4 | - \ ln | x + 4 | \ big] _1 ^ 3 = 3 + 0- \ ln 7-1- \ ln 3+ \ ln 5 = 2 + \ ln \ frac {5} {21} $

Task 5

Calculate
  1. $ \ D \ int \ frac {\ sin x} {1+ \ cos ^ 2 x} \, dx $

  2. $ \ D \ int \ frac {6x ^ 3 + 4x ^ 2-4x-16} {x ^ 4-3x ^ 2-4} \, dx $.
The use of an integral board is not permitted for this task.
tip
  1. Skilfully substitute.
  2. Partial fraction decomposition.

solution
  1. $ \ D t = \ cos x, \ quad \ frac {dt} {dx} = - \ sin x \ \ D \ RA \ int \ frac {\ sin x} {1+ \ cos ^ 2 x} \, dx = \ int \ frac {\ sin x} {1 + t ^ 2} \ cdot \ left (- \ frac {1} {\ sin x} \ right) \, dt \ \ D = - \ int \ frac {1} {1 + t ^ 2} \, dt = - \ arctan (t) + C = - \ arctan {(\ cos x)} + C $
  2. Partial fraction decomposition:

    $ \ D x ^ 4-3x ^ 2-4 = (x ^ 2 + 1) (x ^ 2-4) = (x ^ 2 + 1) (x-2) (x + 2) \ \ D \ RA f (x) = \ frac {6x ^ 3 + 4x ^ 2-4x-16} {x ^ 4-3x ^ 2-4} = \ frac {Ax + B} {x ^ 2 + 1} + \ frac {C} {x-2} + \ frac {D} {x + 2} \ \ D \ LR 6x ^ 3 + 4x ^ 2-4x-16 = (Ax + B) (x ^ 2-4) + C (x ^ 2 + 1) (x + 2) + D (x ^ 2 + 1) (x-2) $

    Substituting $ x = 2 $ and $ x = -2 $:

    $ \ D \ quad 48 + 16-8-16 = \ quad 20C \ LR C = 2 \ \ D-48 + 16 + 8-16 = -20D \ LR D = 2 $

    Compare the terms with $ x ^ 3 $ and $ x ^ 2 $:

    $ \ D 6x ^ 3 = x ^ 3 \ left (A + 2 + 2 \ right) \ LR A = 2 \ \ D 4x ^ 2 = x ^ 2 \ left (B + 4-4 \ right) \ LR B = 4 \ \ D \ RA f (x) = \ frac {2x + 4} {x ^ 2 + 1} + \ frac {2} {x-2} + \ frac {2} {x + 2} \ \ D \ int f (x) \, dx = 2 \ left (\ ln \ left | x-2 \ right | + \ ln \ left | x + 2 \ right | \ right) + \ int \ frac { 2x} {x ^ 2 + 1} + \ frac {4} {x ^ 2 + 1} \, dx \ \ D = 2 \ ln \ left | x ^ 2-4 \ right | +4 \ arctan x + \ int \ frac {2x} {x ^ 2 + 1} \, dx \ \ D \ quad u = x ^ 2, \ quad \ frac {du} {dx} = 2x \ \ D = 2 \ ln \ left | x ^ 2-4 \ right | +4 \ arctan x + \ int \ frac {1} {u + 1} \, du \ \ D = 2 \ ln \ left | x ^ 2-4 \ right | +4 \ arctan x + \ ln \ left | u + 1 \ right | + c \ \ D = 2 \ ln \ left | x ^ 2-4 \ right | +4 \ arctan x + \ ln \ left (x ^ 2 + 1 \ right) + c $

Exercise 6

Let $ f \ in C ^ 1 [0,2 \ pi] $ and $ \ D b_k = \ int_0 ^ {2 \ pi} f (x) \ cos kx \, dx $.
Show $ \ D \ lim_ {k \ to \ infty} b_k = 0 $ using partial integration.
tip

$ \ D \ int_a ^ b u '(x) \ cdot v (x) \, dx = u (x) \ cdot v (x) \ big | _a ^ b- \ int_a ^ bu (x) \ cdot v' (x) \, dx $


solution
$ \ D b_k = f (x) \ sin (kx) \ frac 1k \ Bigg | _0 ^ {2 \ pi} - \ int_0 ^ {2 \ pi} f '(x) \ sin (kx) \ frac 1k \ , dx \ \ D = \ frac 1k \ left (f (2 \ pi) \ sin (2 \ pi k) -f (0) \ sin (0) \ right) - \ frac 1k \ int_0 ^ {2 \ pi} f '(x) \ sin (kx) \, dx = - \ frac 1k \ int_0 ^ {2 \ pi} f' (x) \ sin (kx) \, dx $

The integral converges because the integrand is continuous and the interval compact, so $ b_k $ tends to zero for $ k \ to \ infty $ because of the factor $ \ frac 1k $.

Exercise 7

Examine $ \ D \ int_0 ^ \ infty \ frac {1} {\ sqrt {1 + e ^ x}} \, dx $ for convergence.
Note: it is not necessary to calculate an antiderivative.
tip

Majorant criterion.


solution
$ \ D 0 \ leq \ frac {1} {\ sqrt {1 + e ^ x}} \ leq \ frac {1} {\ sqrt {e ^ x}} = e ^ {- \ frac x2} \ \ D \ int_0 ^ \ infty e ^ {- \ frac x2} \, dx = -2e ^ {- \ frac x2} \ Big | _0 ^ \ infty = 2 \ \ RA 0 \ leq \ int_0 ^ \ infty \ frac {1} {\ sqrt {1 + e ^ x}} \, dx \ leq 2 $

The integral is not negative and is bounded upwards by the majorant $ 2 $, i.e. it is convergent.

Exercise 8

Let $ \ D G (a) = \ int_1 ^ a \ frac {\ ln (1 + a ^ 2t ^ 2)} {t ^ 2} \, dt $. Calculate $ G '(a) $. What is $ G '(1) $?
tip

$ \ D \ int \ frac {1} {1 + x ^ 2} \, dx = \ arctan x + c $


solution

$ \ D G '(a) = \ frac {\ ln (1 + a ^ 4)} {a ^ 2} + \ int_1 ^ a \ frac {1} {t ^ 2} \ frac {1} {1+ a ^ 2 t ^ 2} 2at ^ 2 \, dt \ \ D = \ frac {\ ln (1 + a ^ 4)} {a ^ 2} + \ int_1 ^ a \ frac {2a} {1 + a ^ 2 t ^ 2} \, dt \ \ D s = at \ LR t = \ frac sa \ RA t = 1 \ LR s = a, \ quad t = a \ LR s = a ^ 2, \ quad \ frac {dt} {ds} = \ frac 1a \ \ D = \ frac {\ ln (1 + a ^ 4)} {a ^ 2} + \ int_a ^ {a ^ 2} \ frac {2a} {1 + s ^ 2} \ frac 1a \, ds \ \ D = \ frac {\ ln (1 + a ^ 4)} {a ^ 2} + \ int_a ^ {a ^ 2} \ frac {1} {1 + s ^ 2} \, ds \ \ D = \ frac {\ ln (1 + a ^ 4)} {a ^ 2} +2 (\ arctan a ^ 2- \ arctan a) \ \ D G ' (1) = \ ln 2 $

Exercise 9

Examine the following improper integrals for convergence:
  1. $ \ D \ int_1 ^ \ infty e ^ {\ frac {1} {x}} \, dx $
  2. $ \ D \ int _ {- 1} ^ 0 e ^ {\ frac {1} {x}} \, dx $
  3. $ \ D \ int_0 ^ \ infty \ frac {\ sin x} {1 + x ^ 3} \, dx $

tip

Find integrable majorante or minorante.


solution
  1. The integrand is monotonically decreasing from $ e ^ 1 $ to $ e ^ 0 = 1 $, therefore $ e ^ \ frac {1} {x} \ ge 1 $ and the integral does not exist.
  2. It is $ \ D \ lim_ {x \ to 0-} e ^ \ frac {1} {x} = 0 $. Therefore the integrand $ f (x) = e ^ \ frac {1} {x} $ can be continued continuously through $ f (0) = 0 $ and the integral exists.
    Alternative:

    $ \ D t = \ frac {1} {x}, \ quad x = \ frac {1} {t}, \ quad \ frac {dx} {dt} = - \ frac {1} {t ^ 2} \ \ \ D \ int _ {- 1} ^ 0 e ^ \ frac {1} {x} \, dx = - \ int _ {- 1} ^ {- \ infty} \ frac {1} {t ^ 2} e ^ t \, dt = \ int _ {- \ infty} ^ {- 1} \ frac {1} {t ^ 2} e ^ t \, dt. $

    Now $ e ^ t $ (or $ \ frac {1} {t ^ 2} $) is an integrable majorante.

  3. It is

    $ \ D \ int_0 ^ \ infty \ frac {\ sin x} {1 + x ^ 3} \, dx = \ int_0 ^ 1 \ frac {\ sin x} {1 + x ^ 3} \, dx + \ int_1 ^ \ infty \ frac {\ sin x} {1 + x ^ 3} \, dx $.

    The first integral exists because continuous functions can be integrated on compact intervals, in the second $ \ frac {1} {x ^ 3} $ is an integrable majorant.

Exercise 10

Let $ \ D f (x) = \ frac {1} {1 + x ^ 2} $. We are looking for an approximation for $ \ D \ int_0 ^ 1f (x) \, dx $.
Only the function values ​​at the positions $ x = 0 $, $ x = \ frac 12 $ and $ x = 1 $ should be used.

  1. What is the value of the summed trapezoidal rule?

  2. Prove that the error is less than $ \ frac {1} {16} $.

  3. What is the value of Kepler's barrel rule?

  4. What is the exact value of the integral?

tip
  1. Summed trapezoidal rule:

    $ \ DT ^ {(n)} (f) = h \ left (\ frac 12 f (a) + \ frac 12 f (b) + \ sum_ {i = 1} ^ {n-1} f (a + ih) \ right), \ quad h = \ frac {ba} {n} $
  2. Remaining term of the summed trapezoidal rule:

    $ \ DE ^ {(n)} (f) = - \ frac {(ba)} {12} h ^ 2 f '' (\ xi), \ quad \ text {with suitable intermediate point} \ xi \ in [a , b] $
  3. Kepler's barrel rule (Simpson rule):

    $ \ D S (f) = \ frac {b-a} {6} \ left (f (a) + 4f (\ frac {a + b} {2} + f (b) \ right). $
  4. $ \ int \ frac {1} {1 + x ^ 2} \, dx = \ arctan (x) + c

solution
$ f (0) = 1, \ quad f (\ frac 12) = \ frac 45, \ quad f (1) = \ frac 12 $
  1. here:

    $ \ D a = 0, \ quad b = 1, \ quad h = \ frac 12, \ quad n = 2 \ \ DT ^ {(2)} (f) = \ frac 12 \ left (\ frac 12 \ left (f (0) + f (1) \ right) + f (\ frac 12) \ right) = \ frac 12 \ left (\ frac 12 \ left (1+ \ frac 12 \ right) + \ frac 45 \ right) = \ frac {31} {40} = 0 {,} 775 $
  2. $ \ D f '' (x) = \ frac {6x ^ 2-2} {\ left (1 + x ^ 2 \ right) ^ 3} \ \ DE ^ {(2)} (f) = - \ frac {1-0} {12} \ frac 14 \ frac {6 \ xi ^ 2-2} {\ left (1+ \ xi ^ 2 \ right) ^ 3} \ leq \ frac {1} {24}<\frac{1}{16} $="">
  3. $ \ D S (f) = \ frac 16 \ left (1 + 4 \ cdot \ frac 45 + \ frac 12 \ right) = \ frac {47} {60} \ approx 0 {,} 783 $
  4. $ \ D \ int_0 ^ 1 \ frac {1} {1 + x ^ 2} \, dx = \ arctan (x) \ Bigg | _0 ^ 1 = \ frac \ pi 4 \ approx 0 {,} 785 $

Exercise 11

Assign an antiderivative to each
  1. $ \ D \ frac {x-2} {x ^ 3 + x ^ 2 + x + 1} $
  2. $ \ D \ frac {1} {\ sqrt {1 + e ^ x}} $.

tip
  1. Partile fraction decomposition.
  2. Use the substitution $ u = \ sqrt {1 + e ^ x} $.

solution
  1. $ x ^ 3 + x ^ 2 + x + 1 = (x + 1) (x ^ 2 + 1) $.

    Partial fraction decomposition:

    \ begin {eqnarray *} \ frac {x-2} {(x + 1) (x ^ 2 + 1)} & = & \ frac {A} {x + 1} + \ frac {Bx + C} {x ^ 2 + 1} \ x-2 & = & A (x ^ 2 + 1) + (Bx + C) (x + 1) \ end {eqnarray *}

    Substituting $ x = -1 $ gives:

    $ \ D-3 = 2A \ LR A = - \ frac {3} {2} $.

    Comparison of the terms at $ x ^ 2 $:

    $ \ D 0 = - \ frac {3} {2} + B \ LR B = \ frac {3} {2} $.

    Substituting $ x = 0 $:

    $ \ D-2 = - \ frac {3} {2} + C \ LR C = - \ frac {1} {2} $.

    This results in

    $ \ D \ int \ frac {x-2} {x ^ 3 + x ^ 2 + x + 1} \, dx \ \ D = \ frac {1} {2} \ int \ left (\ frac {- 3} {x + 1} + \ frac {3x} {x ^ 2 + 1} - \ frac {1} {x ^ 2 + 1} \ right) \, dx \ \ D = - \ frac {3} {2} \ ln | x + 1 | + \ frac {3} {4} \ ln (x ^ 2 + 1) - \ frac {1} {2} \ arctan x + c $

  2. $ \ D u = \ sqrt {1 + e ^ x} \ LR x = \ ln (u ^ 2-1), \ quad \ frac {dx} {du} = \ frac {2u} {u ^ 2-1 } \ \ D \ int \ frac {1} {\ sqrt {1 + e ^ x}} \, dx = \ int \ frac {1} {u} \ frac {2u} {u ^ 2-1} \ , du = \ int \ Big (\ frac {1} {u-1} - \ frac {1} {u + 1} \ Big) \, du \ \ D = \ ln | u-1 | - \ ln | u + 1 | + c = \ ln | \ sqrt {1 + e ^ x} -1 | - \ ln | \ sqrt {1 + e ^ x} +1 | + c \ $

    In the approach for the partial fraction decomposition

    $ \ D \ frac {2} {u ^ 2-1} = \ frac {A} {u-1} + \ frac {B} {u + 1} \ LR 2 = A (u + 1) + B ( u-1) $

    $ u = 1 $ and $ u = -1 $ are used.