Integral - Flächenberechnungen

Berechne die Integrale der folgenden Funktionen im angegebenen Interval

a) f(x) = 2x  [1, 3 ]

³∫₁ (2x) =
³∫₁ (2/2x²) = 3² – 1² = 8


b) f(x) = x/2 + 1 [-2, 2]


²∫_-2 (x/2 + 1) =
²∫_-2 (x²/4 + x) = (4/4 + 2) – (4/4 – 2) = 1 +2 – 1 + 2 = 4


d) f(x) = x² [1,3]


³∫₁ ( x² ) =
³∫₁ (x³/3) = 27/3 – 1/3 = 26/3


e) f(x) = x²/4 + 2  [0, 4]


⁴∫₀ ( x³/4 * 3 + 2x) = 64/12 + 8 = 64/12 + 96/12 = 160/12 = 13,33


j) f(x) = x³/4 - 3x²/2 + 7x/2   [ 0,3 ]


³∫₀ ( x³/4 - 3x²/2 + 7x/2 ) =
³∫₀ ( x⁴/4*4 - 3x³/2*3 + 7x²/2*2 ) = 81/16 – 81/6 + 63/4 = 243/48 - 648/48 + 756/48 = 351/48 = 117/16 ~ 7,31



2.) Berechne den Inhalt der Fläche zwischen Kurve und x-Achse:

d) f(x) = x³ - 6x² + 9x

x(x² - 6x + 9)

x = 0

x² - 6x + 9 = 0 

kl. Lösungsformel: 
6/2 +- √(-6/2)² – 9 => 3

 
Nullstellen: 0,3

³∫₀ (x³ - 6x² + 9x)
³∫₀ (x⁴/4 – 6x³/3 + 9x²/2) = 81/4 – 162/3 + 81/2 = 243/12 - 648/12 + 486/12 = 81/12 ~6,75


f) f(x) = x³ - 8x² + 15x
x³ - 8x² + 15x
x (x² -8x +15)
x = 0


kl Lösungsformel
8/2 +- √(-8/2)² – 15
4 +- 1 = > 5, 3


x1 = 0, x2 = 5, x3 = 3,


³∫₀ (x³ - 8x² + 15x ) =
³∫₀ (x⁴/4 – 8x³/3 + 15x²/2) =
81/4 – 216/3 + 135/2 = 243/12 – 864/12 + 810/12 = 189/12


⁵∫₃ (x³ - 8x² + 15x ) =
³∫₀ (x⁴/4 – 8x³/3 + 15x²/2) =
625/4 – 1000/3 + 375/2 – (81/4 – 216/3 + 135/2) =
1875/12 - 4000/12 + 2250/12 - 243/12 + 864/12 - 810/12 = |-64/12 | = 64/12


A = 189/12 + 64/12 = 253/12 ~ 21,08


g) f(x) = x³/3 - 3x


x(x²/3-3)
x = 0
x²/3 = 3
x² = 9 / √
x = +- 3 x1= 0, x2= 3, x3 = -3


⁰∫_-3 (x³/3 - 3x )
⁰∫_-3 (x⁴/3*4 – 3x²/2 ) =
-81/12 + 27/2 = 81/12


³∫₀ (x³/3 - 3x )
³∫₀ (x⁴/3*4 – 3x²/2 ) = 81/12 - 27/2 = 81/12 - 162/12 = | - 81/12 | = 81/12


A = 81/12 + 81/12 = 162/12 = 13,5



h) f (x) = x⁴ - 5x² + 4


u = x² u² +5u + 4


5/2 +- √(5/2)² – 4
2,5 +- 1,5
u = 4, 1
=> 4 => +- 2
=> 1 => +- 1


^-1∫_-2 (x⁴ - 5x² + 4)
^-1∫_-2 (x⁵/5 – 5x³/3 + 4x)
-1/5 + 5/3 – 4 - (-32/5 + 40/3 – 8) = -3/15 + 25/15 - 60/15 + 96/15 - 200/15 + 120/15
= | -22/15 |= 22/15


¹∫_-1 (x⁴ - 5x² + 4) =
¹∫_-1 (x⁵/5 – 5x³/3 + 4x) =
1/5 - 5/3 + 4 - (-1/5 + 5/3 – 4 ) = 3/15 - 25/15 + 60/15 + 3/15 - 25/15 + 60/15 = 76/15


²∫₁ (x⁴ - 5x² + 4) =
²∫₁ (x⁵/5 – 5x³/3 + 4x) =
32/5 - 40/3 + 8 - 1/5 - 5/3 + 4 =
96/15 - 200/15 + 120/15 - 3/15 + 25/15 – 60/15 = -22/15 = | -22/15 |

A = 22/15 + 76/15 + 22/15 = 120/15 = 8

3. Berechne den Inhalt der Fläche zwischen den beiden Kurven:

a) f(x) = x², g(x) = x + 6

x² = x + 6

x² – x – 6 = 0

½ +- √(-1/2)² + 6

0,5 +- 2,5

=> x1 = 3, x2 = -2

³∫₀ (x² – x – 6 ) = x³/3 – x²/2 – 6x =
27/3 – 9/2 – 18 + 8/3 + 4/2 – 12 =
35/3 – 5/2 – 30
70/6 – 15/6 – 180/6 = |-125/6| = 20,83

b) f(x) = 4x - x², g(x) = x

4x – x² = x

4x – x² – x = 0

3x – x² = 0

x * (3 – x)

=> x1 = 0, x2 = 3

³∫₀ (3x – x²) =
³∫₀ 3x²/2 – x³/3 =
27/2 -27/3 = 81/6 – 54/6 = 27/6 = 4,5


c) f(x) = x², g(x) = 4x - x²

x² = 4x – x²

4x – 2x² = 0

x( 4 – 2x) = 0

=> x1 = 0, x2 = 4 -2x = 2

³∫₀ (4x – 2x²) = ²∫₀ (4x²/2 – 2x³/3) = 16/2 - 16/3 = 48/6 – 32/6 = 16/6 = 2,67


d) f(x) = x², g(x) = 5 – x²/4


x² = 5 – x²/4
4x² / 4 = 5 – x²/4 / + x²/4
5x² / 4 = 5 / *4
5x² = 20 /:5
x² = 4 / √


x = +- 2

²∫_-2 (5x²/4 – 5) =

²∫_-2 (5x³/12 – 5x) =

40/12 – 10 - (- 40/12 + 10) =

40/12 – 10 + 40/12 - 10 =

40/12 – 120/12 + 40/12 – 120/12 = -160/12 = | -160/12 | = 13.33