高等数学 第6小题和第四大题 过程纸上 详细
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6. F(x) = ∫ <0, x>(x^2-t^2)f''(t)dt
= x^2∫ <0, x>f''(t)dt - ∫ <0, x>t^2f''(t)dt
F'(x) = 2x∫ <0, x>f''(t)dt + x^2f''(x) - x^2f''(x)
= 2x∫ <0, x>f''(t)dt
lim<x→0>F'(x)/x^2
= lim<x→0> 2∫<0, x>f''(t)dt / x (0/0)
= lim<x→0> 2f''(x) / 1 = 1,
则 f''(0) = 1/2
四。 I(a) = ∫<0, π>(asinx - π)^2dx
= ∫<0, π>[a^2(sinx)^2 - 2aπsinx + π^2]dx
= ∫<0, π>[ π^2 + a^2/2 - (1/2)a^2cos2x - 2aπsinx ]dx
= [( π^2 + a^2/2)x - (1/4)a^2sin2x +2aπcosx]<0, π>
= ( π^2 + a^2/2)π -4aπ
= (π/2)(a^2 - 8a + 2π^2)
dI(a)/da = (π/2)(2a - 8),
则 a = 4 时 I(a) 最小.
= x^2∫ <0, x>f''(t)dt - ∫ <0, x>t^2f''(t)dt
F'(x) = 2x∫ <0, x>f''(t)dt + x^2f''(x) - x^2f''(x)
= 2x∫ <0, x>f''(t)dt
lim<x→0>F'(x)/x^2
= lim<x→0> 2∫<0, x>f''(t)dt / x (0/0)
= lim<x→0> 2f''(x) / 1 = 1,
则 f''(0) = 1/2
四。 I(a) = ∫<0, π>(asinx - π)^2dx
= ∫<0, π>[a^2(sinx)^2 - 2aπsinx + π^2]dx
= ∫<0, π>[ π^2 + a^2/2 - (1/2)a^2cos2x - 2aπsinx ]dx
= [( π^2 + a^2/2)x - (1/4)a^2sin2x +2aπcosx]<0, π>
= ( π^2 + a^2/2)π -4aπ
= (π/2)(a^2 - 8a + 2π^2)
dI(a)/da = (π/2)(2a - 8),
则 a = 4 时 I(a) 最小.
追问
倒数第四行是减去2aπ吧
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