第1个回答 2011-12-15
答案是3/4
∫[0，x] t * f(2x-t) dt = 0.5 * arctan(x²)
代换：y = 2x-t，dy = -dt
t = 0，y = 2x；t = x，y = x
∫[2x，x] (2x-y) * f(y) (-dy) = 0.5 * arctan(x²)
(2x)∫[x，2x] f(y) dy - ∫[x，2x] yf(y) dy = 0.5 * arctan(x²)，下一步两边求导
d/dx ∫[x，2x] f(y) dy = (2x)'f(2x) - (x)'f(x) = 2f(2x) - f(x)
d/dx (2x)∫[x，2x] f(y) dy = 2∫[x，2x] f(y) dy + 4xf(2x) - 2xf(x)
d/dx ∫[x，2x] yf(y) dy = (2x)'(2x)f(2x) - (x)'xf(x) = 4xf(2x) - xf(x)
原本的方程变为：
{2∫[x，2x] f(y) dy + 4xf(2x) - 2xf(x)} - {4xf(2x) - xf(x)} = 0.5 * 1/(1+x⁴) * 2x
2∫[x，2x] f(y) dy - xf(x) = x/(1+x⁴)
∫[x，2x] f(y) dy = x/[2(1+x⁴)] + (1/2)xf(x)，下一步将x=1代入，得
∫[1，2] f(x) dx = 1/4 + (1/2)(1)f(1) = 1/4 + (1/2) = 3/4