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x 0 f(t) sin(x−t)dt ． ，f(x) 求 ． 2 (888) x >0 対 ，f(x

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熊本大学 数理科学総合教育

§9 微分積分学 基本定理 不定積分 演習問題 2

問題 難易度 目安 【基礎】899 【標準】889 【発展】888

1 (888) x >−1 定義 関数f(x) x >−1 2回微分可能 ， x >−1

対

f(x) = 1

(x+ 1)2 +

∫ x 0

f(t) sin(x−t)dt

． ，f(x) 求 ．

2 (888) x >0 対 ，f(x) := sinx

x ． 非負整数n 対

f(n)(x) = 1 xn+1

∫ x 0

tncos (

t+ nπ 2

) dt

数学的帰納法 用 証明 ． ，f(n)(x) f(x) 第n階導関数 表 ．

， 用 lim

n→∞f(n)(1) 求 ．

1

参照

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