y = ∫[0,x] t f(x² - t²) dt
令u = x² - t²,du = -2t dt
當t = 0,u = x²；當t = x,u = 0
y = ∫[x²,0] t f(u) * du/(-2t)
= 1/2 ∫[0,x²] f(u) du
dy/dx = 1/2 [2x * f(x²) - 0]
= x f(x²)
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樓上的方法也不錯.
令z² = x² - t²,2z dz = -2t dt => dt = -z/t dz
當t = 0,z = x；當t = x,z = 0
∫[0,x] t f(x² - t²) dt
= ∫[x,0] t f(z²) * (-z/t) dz
= ∫[x,0] -z f(z²) dz
= ∫[0,x] z f(z²) dz
dy/dx = x f(x²) - 0
= x f(x²)
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兩個方法也行,但樓上的替換怎么又會涉及因變量y呢?