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ÇÖZÜMLÜ SORULAR Belirli İntegral ve Uygulamaları MATEMATİK

1 1 1 4 1 4 23. Aşağıda gerçek sayılar kümesinde tanımlı f fonksiyonunun
2
21. 1 ∫∫ (x + x )dx − 2 2 ∫ ∫ (x + x )dx + 2 2 ∫ ∫ (x + x )dx ifadesi aşağıdaki- grafiği verilmiştir.
− − 1 − 1 − 1
y
lerden hangisine eşittir?
145 155 155 175 175 4
A) B) C) D) E)
3 3 6 3 6 3

Çözüm: 3 x
O 1 2
1 1 1 (x + 2 2 2 1 4 4 4 (x + 2 2 2 4 4 4 x )dx = y = f(x)
(x + x )dx =
2 ∫
− 1 ∫ − 1 ∫ (x + − 1 ∫ (x + x )dx − x )dx − x )dx 2 ∫ (x + 2 ∫ 2 ∫ (x + − x )dx + x )dx + x )dx + 2 ∫ (x +
2 ∫ (x + x )dx =
2 2 2
x )dx
− 1 ∫ − 1 ∫ − 1 1 1 1 ∫ (x + x )dx + x )dx + x )dx 1 ∫ (x + 1 ∫ 1 ∫ (x + + 2 2 2 (x + x )dx + x )dx + x )dx + 2 2 2 2 ∫ (x + 2 ∫ (x + 2 ∫ 4 4 4 (x + (x + (x + x )dx –3
2 2 2
2 2 2
x )dx
−
4 4 4
4  − 
−
( 1)   
= = = = 4 − 1 1 ∫ 4 4 4 1 ∫ ∫ ∫ (x + x) dx = dx = x)    x x x 2 2 2 + + +    x x x    =  =         4 4 4 2 2 2 +   = 4 4  (− ( 1) 1) 1) 2 2 2 + + (− + ( 1) 1) 3 3 3    3 3 f(x) f (x) x ⋅ ′
(−
3 3 3
3 3 3
−

− +
(x +
 −
− +

2 2 2
(x + x) dx =




− − 1 −       2 2 2 3 3 3    −        2 2 2 2 2 2  2 2 2 2 2 2       Buna göre ∫ 1 ∫ x 2 dx

− − 1 1 1 1
64 
1   
1 175
   64  1 1 1 1 1 64 64 1 1 1 175
64 
64 1 1
= = =    8 + 8 +  8 +   − − 3    = =  = 8 + 8 + 8 + − − + + − = = = + 175 ifadesinin değeri kaçtır?
− 
−  − − 


2 2
   3 3 3  2 3 3     3 3 3 2 2 2 3 3 3 6 6 6

A) 1 B) 2 C) 3 D) 4 E) 5
Cevap: E Çözüm:
y
4
3
3 x
O 1 2
y = f(x)
–3
−
3 f(x) f (x) x ⋅ ′ 3  f(x)  − ′

∫ 2 dx =    ∫  dx
1 x 1 x 
 f(x)  − 3  f(3) f(1)   − 3 3 
=    =  −  −   = −   −    = 4
 x 1   3 1   3 1 
Cevap: D
22. 3 f(x)dx =
− 2 ∫ 8
3 x f (x)dx = ′ ⋅
− 2 ∫ 10
integralleri veriliyor.
f(3) = 12 olduğuna göre f(–2) değeri kaçtır?
A) –6 B) –8 C) –9 D) –10 E) –12
Çözüm:
Verilen eşitlikler taraf tarafa toplandığında
3 3 3 f(x)dx + 3 3 3 3 ′ ⋅
3
2 ∫ f(x)dx + +
2 ∫ x f (x)dx = x f (x)dx =
x f (x)dx = ′ ⋅
f(x)dx
1818
− ∫ − 2 ∫ ∫ f(x)dx + − ∫ − 2 ∫ ∫ x f (x)dx = 2 ⋅ ′ ′ ⋅ 18
18
− 2 − 2 − 2 −
3 3 3
3
3 3 3 (f(x) x f (x))dx = 3 3 3 3 ′ ⋅ 3
⋅ ⋅ ′
⋅ (x f(x)) dx(x ⋅
′
⋅
(x f(x)) dx(x f(x))
=
+ + ⋅
2 ∫ (x f(x)) dx(x f(x)) f(x))
⋅ f (x))dx
− ∫ − 2 ∫ ∫ (f(x) x f (x))dx =⋅ 2 ∫ (f(x) x f (x))dx = =⋅ − 2 (f(x) + ′ ′ ′ − ∫ − 2 ∫ ∫ (x f(x)) dx(x f(x))
′ x
′
⋅
+
2 ⋅
−
−
−
2
2
2
− − − 2 2 − 2
⋅ ⋅
(− (−f 2(−
) = ⋅
= =
= 3 ⋅ 3 f ⋅ (3 ( ) 2 f 2) = ⋅ 3 12 2 (f ⋅− ( f 2 =
( f 3) −− 2 f 2) = ⋅ 3 12 2) −−
3 ⋅ ( f 3) −−( ) 2 ⋅
+ + ⋅−
) 18
+ ⋅− )2 =)
3 12 2 (f 2 = 1818
3 ⋅
=
( f 3) −−
( ) 2 f 2(−
+ ⋅−
) = ⋅
( ) ⋅
3 12 2 ( f 2 =
) 18
f − f − f − ( ) − − − − 9 9 9
( ) 2 = 2 =( ) 2 =
−
f
( ) 2 =
9
Cevap: C
477