题目
(1992数三)求连续函数f(x)，使它满足int_(0)^1f(tx)dt=f(x)+xsin x.
(1992数三)求连续函数f(x)，使它满足$\int_{0}^{1}f(tx)dt=f(x)+x\sin x$.
题目解答
答案
令 $ u = tx $，则原方程变为
\[
\frac{1}{x} \int_0^x f(u) \, du = f(x) + x \sin x.
\]
两边乘以 $ x $ 得
\[
\int_0^x f(u) \, du = x f(x) + x^2 \sin x.
\]
求导得
\[
f(x) = f(x) + x f'(x) + 2x \sin x + x^2 \cos x,
\]
化简得
\[
f'(x) = -2 \sin x - x \cos x.
\]
积分得
\[
f(x) = \int (-2 \sin x - x \cos x) \, dx = 2 \cos x - (x \sin x + \cos x) + C = \cos x - x \sin x + C.
\]
**答案：**
\[
\boxed{\cos x - x \sin x + C}
\]