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f ( x )= x − x +231 43 (3) f ( x )= x − xf ( t ) dt +2 f ( t ) dt ∫ ∫ f ( x )=2 x +103 x +1 f ( x )= x +23 x (1) f ( x )=2 x +1+ x f ( t ) dt (2) f ( x )= x + xf ( t ) dt ∫ ∫ 1 f ( x ) f ( x ) dx f ( t ) dt A ∫ ∫

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シェア "f ( x )= x − x +231 43 (3) f ( x )= x − xf ( t ) dt +2 f ( t ) dt ∫ ∫ f ( x )=2 x +103 x +1 f ( x )= x +23 x (1) f ( x )=2 x +1+ x f ( t ) dt (2) f ( x )= x + xf ( t ) dt ∫ ∫ 1 f ( x ) f ( x ) dx f ( t ) dt A ∫ ∫"

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(1)

定積分

b a

f (x)dx

や

b a

f (t)dt

は定数である。Aとおく！  発展編

1

 次の等式を満たす関数

f (x)

を求めよ。

(1) f (x) = 2x

2

+ 1 + x

∫

1 0

f(t)dt (2) f (x) = x

2

+

∫

1 0

xf(t)dt

【解答】

f (x) = 2x

2

+ 10

3 x + 1

【解答】

f (x) = x

2

+ 2

3 x (3) f (x) = x

2

−

∫

2 0

xf (t)dt + 2

∫

1 0

f (t)dt

【解答】

f (x) = x

2

− 4 3 x + 2

3

(2)

(4) f (x) =

∫

1 0

xf(t)dt +

∫

1 0

tf(t)dt + 1

【解答】

f (x) = − 12x − 6 (5) f (x) = 1 +

∫

1 0

(x − t)f (t)dt

【解答】

f (x) = 12 13 x + 6

13

2

(3)

a

が定数のとき，

d dx

∫

x a

f (t)dt = f (x)

 すなわち，

(∫

x a

f (t)dt )

′

= f (x)

2

 次の等式を満たす関数

f (x)

と定数

a

の値を求めよ。

(1)

∫

x 1

f (t)dt = 2x

2

+ x + a (2)

∫

a x

f(t)dt = x

3

− 3x

【解答】

f (x) = 4x + 1， a = − 3

【解答】

f (x) = − 3x

2

+ 3， a = 0, ± √ 3

3

 次の関数の極値を求めよ。

(1) f (x) =

∫

x 0

(3t

2

− 2t − 1)dt

【解答】

x = − 1

3

のとき 極大値

5

27

，

x = 1

のとき極小値

− 1

(4)

(2) f (x) =

∫

1 x

(t

3

− t

2

− 2t)dt

【解答】

x = − 1

のとき 極大値

− 2

3

，

x = 2

のとき 極大値

19

12

，

x = 0

のとき極小値

− 13 12

4

 実数係数の多項式

f (x)

と

g(x)

は次の関係式を満たすとする。このとき，f

(x)

と

g(x)

を求めよ。

f (x) = x −

∫

2

−1

g(t)dt， g(x) = 3 + 2

∫

x 0

f (t)dt

【解答】

f (x) = x − 3，g(x) = x

2

− 6x + 3

4

(5)
(6)
(7)
(8)

参照

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